LIBRARY OF CONGRESS, 3®^ Chap.y.i...*edpyfight No Shelf.„i.V_fe_2- •._ UNITED STATES OF AMERICA. I I ELEMENTS OF THE DIFFERENTIAL AND INTEGRAL CALCULUS Witb ^samples an& practical Bpplications / BY J. W. NICHOLSON, A.M., LL.D. President and Professor of Mathematics Louisiana Slate University and Agricultural and Mechanical College <£ FER 13 J 896 . I NEW YORK AND NEW ORLEANS UNIVERSITY PUBLISHING COMPANY 1896 K ^' Copyright, 1896, BY UNIVERSITY PUBLISHING CO. All Rights Reserved ***1673 PEEFACE. Ik many respects this work is quite different from any other on the same subject, though in preparing it there has been no attempt at originality beyond presenting the principles in a more tangible form than usual, and thus securing a better text-book for the ordinary student of mathematics. The aim has been to prepare a work for beginners, and at the same time to make it sufficiently comprehensive for the requirements of the usual undergraduate course. The chief distinction of the treatise is that it is based on the conception of Proportional Variations. This method has been employed as the most elementary and practical, and none the less rigorous or general, form of presenting the first principles of the subject (see the following Note). Differentiation and Integration are carried on together, and the early introduction of practical applications both of the dif- ferential and integral calculus, which this mode of presenting the subject permits, is intended to serve an important purpose in illustrating the utility and potentiality of the science, and arous- ing the interest of the student. The formulas for differentiating and integrating are, as a rule, expressed in terms of u and v instead of x, u and v being functions of x. The advantages thus secured are obvious. Among the additional features of special interest may be mentioned the following : (1) The treatment of dx as a variable independent of x (Art. 68, aud Appendix, A 3 ): (2) a rigorous deduction of a simple test of absolute convergency, without recourse to the remainder in Taylor's formula (Arts. 115 to 119) ; (3) an extension of the ordinary rules for finding maxima and minima (Arts. 140 to 143) ; (4) a chapter on Independent Integration (Chap. IX); (5) integration by indeterminate co- IV PREFACE. efficients (Arts. 211 to 216); (6) the introduction of turns in curve-tracing (Arts. 175 to 179) ; and (7) a new proof of Taylor's formula, which is believed to be as rigorous as, and less artificial than, those in general use (Appendix, A 5 ). In preparing the book the best available authors have been consulted, and many of the examples have been taken from the works of Todhunter, Williamson, Conrtenay, Byerly, Rice and Johnson, Taylor, Osborne, Loomis, and Bowser. I improve this opportunity to tender my thanks to Prof. William Hoover of the Ohio University, Prof. Alfred Hume of the University of Mississippi, and Prof. 0. D. Smith of the Polytechnic Institute of Alabama for valuable assistance in the reading of proofs. Their corrections and suggestions have relieved the treatise from various imperfections it would other- wise have contained. Further acknowledgments of indebtedness are also due to my colleague, Prof. C. Alphonso Smith, of the Department of English, who has aided me with his scholarly criticisms. James W. Nicholson. Baton Rouge, La., 1896. Note. — The method of Proportional Variations, which is the suggestion and outgrowth of work in the class-room, is believed to possess the follow- ing merits : (a) The conception is one with which the student is already familiar, for the principle of proportional changes is among the first that he encoun- ters, even in the lower mathematics. (b) It affords finite differentials, and, without introducing infinitesimals. or infinitely small quantities, or "the foreign element of time," has all the advantages of the differential notation. (c) In many cases the proportional variations (or differential) can be detected by inspection (see Arts. 31, 32, 35), and in all cases they may be deduced by the theory of limits. Hence the method has all the lucidity of finite differences and all the rigor of the doctrine of limits. (d) It is a method to which the doctrines of Infinitesimals and Kate of Change are easy corollaries. (<) In general, the form and properties of the increments of all quan- tities are due to proportion and acceleration, or to proportional and dispro- portional changes ; hence, a system of Calculus based on such changes adapts itself naturally to questions in Geometry, Mechanics, and Physics. CONTENTS. CHAPTER I. FUNDAMENTAL PRINCIPLES. [Pages 1 to 20.] Quantity. Variables and constants, Art. 1.* Dependent and indepen- dent variables, 2. Functions, 3. Increasing and decreasing functions, 4. Explicit and implicit functions, 5. Algebraic and Transcendental func- tions, 6. Continuous functions, 7. Notation of functions, 8. Examples. Increments. Increments of independent variable and function, nota. tion, and illustrations, 9. General formula, 10. Examples. Variation. Proportional variation, 11. Principles, 12 to 15. Dispro- portional variation, 16. Principles, 17 to 19. Composition of increments, illustrations, 20. Theory of Limits. Definition and illustrations of limits,, 21. Prin- ciples, 22, 23. Proof of the formula, Ay = mji -f- m 128. Examples. To find the value of 7t, 129. To compute natural loga- rithms, 130. To compute common logarithms, 131 . Indeterminate Forms. How the form — arises, 132. To evaluate functions of the form — , 133. Examples. Of the form — , 134. Examples. Of the forms X oo and oo — oo , 135. Examples. Of the forms 0°, oo °, and 1 , 136. Examples. Implicit functions, 137. Examples. CHAPTER VI. MAXIMA AND MINIMA. [Pages 106 to 121.] Definitions and Principles, Definitions and illustrations, 138. If J[a') is a max. or min. then f'(a') = or oo, 139. f(a') is neither a max. Vlll CONTENTS. noramin., if an even number of the roots off'(x) = and f(x) = go are equal to a', 140. f(a') is a max. or min. if an odd number of the roots of f[x) = and f(x) = go are equal to a', 141. Max. and min. occur alternately, 142. Rules for Finding Maxima and Minima. When all the roots of f(x) = and go are known or can be conveniently found, 143. I. By sub- stituting a' — li and a' -]- Mor x, 144. II. By Taylor's formula, 145. Self- evident principles which serve to facilitate the solution of problems, 146. Examples and problems. Functions of two Independent Variables. Definition, 147. Con- ditions for maxima and minima, 148. Examples. CHAPTER VII. APPLICATIONS OF THE DIFFERENTIAL CALCULUS TO PLANE CURVES. [Pages 122 to 159.] Tangents, Normals, and Asymptotes. Equations of the tangent and normal, 149. Lengths of tangent, normal, subtangent, and subnormal, 150. Examples. Lengths of tangent, normal, subtangent, and subnormal in polar co-ordinates, 151. Examples. Asymptote, 152. General equa- tion of an asymptote, 153. Relation of y to x when they are infinite, 154. Examples. Asymptotes determined by inspection, 155. Examples. Curvature. Total curvature, 156. Uniform curvature, 157. Variable curvature, 158. Radius of curvature, 159. Examples. Radius of curva- ture in polar co-ordinates, 160. Examples. Contact of Different Orders. Definitions, 161. When two curves cross or do not cross at their point of contact, 162. Examples. Osculating curves, 163. Osculating straight line, 164. Osculating circle, 165. Involutes and Evolutes. Definitions, 166. Elementary Principles, 167. To find the equation of the evoluteof a given curve, 168. Examples. Envelopes. Definition, 169. The envelope is tangent to every curve of the series, 170. To find the equation of the envelope of a given series of curves, 171. Examples. Tracing 1 Curves. The general form of a curve, etc., 172. Direction of curvature, convex and concave arcs, 173. Point of inflection and prin- ciples, 174. Examples. Singular Points. Definition (etc.). a'-turns and ?/-turns, and multiple points, 175. To determine the positions of the singular points of curves, 176. To determine the character of the multiple points of curves, 177. Examples. Tracing polar curves, 178. Examples. The character of multiple points often more easily determined by changing to polar co ordi- nates, 179. Examples. CONTENTS. IX CHAPTER VIII. GENERAL DEPENDENT INTEGRATION. [Pages 160 to 190.] Fundamental Formulas. Twenty-two formulas, 180. Reduction and Integration of Differentials. Reduction of Differ- entials, definition, and bow effected, 181. By constant multipliers, 182. Examples. Reduction of Differentials by Decomposition. How effected, 183. Elementary differentials, 184. Examples. Trigonometric differentials, 185. Examples. How trigonometric differentials may often be more conveniently integrated, 186. Rational fractions, 187. When the simple factors of the denominator are real and unequal, 188. Examples. Wben some of tbe simple factors of tbe denominator are real and equal, 189. Examples. Wben some of the factors of tbe denominator are imaginary and unequal, 190. Examples. Wben some of tbe simple factors of the denominator are imaginary and equal, 191. Example. Reduction and Integration by Substitution. Irrational differen- tials, 192. Examples. Wben a -f- bx is tbe only part having a fractional exponent, 193. Examples. Wben \/a -\- bx -\- x- or \/a -\- bx — x* is tbe only sard involved, 194. Examples. Binomial differentials, 195. Con- ditions of integrability, 196. Examples. Integration by Parts. Fundamental formula, 197. Examples. Reduction Formulas. Definition, 198. Reduction formula fc* xP log" xdx, 199. Examples. Reduction formula for a x x n dx, 200. Examples. Reduction formulas for x n cosaxdx and x n sin axdx, 201. Examples. Reduction formula for X sin -1 xdx, 202. Examples. d t, fly* cl "V Integral of , " ,-203. Integral of -- and , 204. a -\-b cos x sin x cos x Approximate Integration. Last resort in separating a differential into its integrable parts, 205. Examples. Development of functions by exact and approximate integration, 206. Examples. CHAPTER IX. INTEGRATION CONTINUED. [Pages 191 to 210.] Independent Integration. Increments deduced from differentials, 207. Examples. Increments as definite integrals, 208. Examples. A more convenient series, 209. Examples. Bernouilli's series, 210. Integration by Indeterminate Coefficients. Explanation and formu- la, 211. When h = or the integration is independent, 212. Illustra- tions and examples. Application to /sin" 1 x cos 71 x dx, 213. When k is not X CONTENTS. = or when the integration is partly dependent, 214. Illustrations and examples. Reduction formulas for binomial differentials, 215. Approx- imate integration and the elliptic differential, 216. CHAPTER X. INTEGRATION AS A SUMMATION OF ELEMENTS. [Pages 211 to 246.] Elements of Functions. Differentials may be as small as we please, 217. Elements, 218. Signification of a definite integral as a sum, 219. Illustrative examples. When dx is not a constant, 220. Signification of a definite integral as the limit of a sum, 221. Illustrative example. Inte- gration equivalent to two distinct operations, 222. Application to Geometry. Length of curves, rectangular co-ordi- nates, 223. Examples. Polar co-ordinates, 224. Examples. To find the equation of a curve when its length is given, 225. Areas of curves, rect- angular co-ordinates, 226. Examples. Generatrix of area, 227. Polar co-ordinates, 228. Examples. Areas of surfaces of revolution, 229. Ex- amples. Volumes of solids of revolution, 230. Successive Integration. A double integral, 231. Definite double integrals, 232. A triple integral, 233. Examples. Areas of Surfaces. Plane surfaces, rectangular co-ordinates, 234. Polar co-ordinates, 235. Examples. Surfaces in general, 236. Examples. Volumes of Solids Determined by Triple Integration. Formula, 237. Examples. Application to Mechanics. Woee, how computed, 238. Example. Centre of Gravity. Centre of gravity, moments, etc., 239. Centre of gravity of a plane area, 240. Centre of gravity of a plane curve, 241. Centres of gravity of solids and surfaces of revolution, 242. Examples APPENDIX. [Pages 247 to 256.] Differentiate functions, Ai. Another illustration of the formula Ay = mji -j- m-tft?, A 2 . The differential of an independent variable is, in general, a variable, A 3 . Another method of finding the differentials of a v and loga v, A 4 . A rigorous proof of Taylor's formula, A 5 . Completion of Maclaurin's formula. A 6 . The values of nil and m 2 in the formula Ay = rriih -f- ra 2 A 2 , A 7 . LIMITED COURSES. (a) The first three chapters. This course is complete as far as it goes, since differentiation and integration are carried on together. It embraces the notation, fundamental principles, and some of the most important applications of the Calculus. The student who understands elementary algebra and geometry, and the construction of elementary loci, should find but little dif- ficulty in mastering it. (b) The first six chapters. This adds to the former course the transcendental functions, development of functions, evalua- tion of the indeterminate forms, and maxima and minima. Suggestions. (1) It is recommended to omit the more difficult examples and problems in passing over the book the first time. (2) A 5 of the Appendix may be substituted for Arts. 116 to 122, at the discretion of the teacher. xi DIFFERENTIAL AND INTEGEAL CALCULUS. CHAPTER I. FUNDAMENTAL PRINCIPLES. QUANTITY. 1. There are two kinds of quantities employed in Calculus, variables and constants. Variables are quantities whose values are to be considered as changing or changeable. They are usually represented by the final letters of the alphabet. Constants are quantities whose values are not to be consid- ered as changing or changeable. They are usually represented by the first letters of the alphabet. Particular values of vari- ables are constants. 2. Dependent and Independent Variables. A Depend- ent Variable is one that depends upon another variable for its value, and an Independent Variable is one that does not depend on another variable, but one to which any arbitrary value or change of value may be assigned. In the elementary differential and integral calculus, the independent variable is usually restricted to real values. Thus, in u = x* — 7x -f 5, v = (1 — x*f, y = log (1 + x), u, v and y are dependent variables, since they depend on the 1 2 DIFFERENTIAL AND INTEGRAL CALCULUS. variable x for their values; but x is au independent variable, since, as we may suppose, any value may be assigned to it without reference to any other variable. 3. Functions. Dependent variables are usually called func- tions of the variables on which they depend. Hence, one vari- able is a function of another when the first depends upon the second for its value, or when the two are so related that changes in the value of the latter produce changes in the value of the former. Thus, the area of a varying square is a function of its side; the cost of cloth is a function of the quality and quantity; the space described by a falling body is a function of the time; every mathematical expression depending on x for its value, as x*, (3x — 7) 3 , 5x 2 — 6x -f 11, etc., is a function of x. 4. Increasing and Decreasing Functions. An Increasing Function is one that increases when the variable increases, as (x -\- 1) 2 , 3x% log (5 -}- x); and a Decreasing Function is one 5 that decreases when the variable increases, as 1/10 — x 2 , -, etc. x A function of x may be increasing for certain values of x, and decreasing for other values. Thus, y = x 2 — 4x -f- 5 is a decreasing function for all values of x < 2, but increasing for all values of x > 2. 5. Explicit and Implicit Functions. An Explicit Function is one whose value is directly expressed in terms of the variable and constants. Thus, in the equations y = (a — x) 5 , y = x* -f Zx 4- 5, y is an explicit function of x. An Implicit Function is one whose value is implied in an equation, but not expressed directly in terms of the variable and constants. Thus, in the equation x 2 -j- 2xy + 5y = 10, y is an implicit function of x, and x is an implicit function of y. By solving the equation for x or y, the function becomes explicit. FUNDAMENTAL PRINCIPLES. 3 6. Algebraic and Transcendental Functions. One variable is called an Algebraic Function of another when the two are connected by an algebraic equation; that is, an equation which contains a finite number of terms involving only constant integral powers of the variables, or an equation which admits of being reduced to this form. Thus, in y = x 2 — 5x, or x 2 y* — xy b -f- 8xy — 5 = 0, or yl — Vax* + y = 7, y is an algebraic function of x, and vice versa. If two variables are connected by an equation which is not algebraic, each is called a Transcendental Function of the other. Thus, if y = sin x, y is a transcendental function of x, and x of y. The following are the elementary transcendental functions: A Logarithmic Function is one that involves the logarithm of a variable; as, log x, log {a -\- y). An Exponential Function is one in which the variable enters as an exponent; as, a x , y x . A Trigonometric Function is the sine, cosine, tangent, etc., of a variable angle ; as, sin x, cos y. An Inverse-Trigonometric Function is an angle whose sine, cosine, tangent, etc., is a variable; as, sin -1 x, cos -1 ?/, tan -1 t, etc., which are read, "an angle whose sine is x" "an angle whose cosine is y" etc, 7. Continuous Functions. A function of a variable is con- tinuous between certain values of the variable (1) when it has a finite value for every value of the variable, and (2) when the changes in its value corresponding to indefinitely small changes in the value of the variable are themselves indefinitely small. Thus, in y = ax -\- b, or y — sin x, or y = e x , y is continu- ous for all finite real values of a:; so also in y = Va? — x 2 , but as a real quantity only for real values of x > — a and < a. x Again, y = is not continuous between the limits x = 1 X rJ and x = 3, for when x =. 2, y = oo . 4 DIFFERENTIAL AND INTEGRAL CALCULUS. 8. Notation of Functions. The symbol f(x) is used to denote any function of x, and is read, " function of x." To de- note different functions of x we employ other symbols, as F(x) y f(x), (p(x), 0(x), etc. According to this notation, y —f(x) rep- resents any equation between x and y when solved for y. Thus, solving the equation y 2, — 2axy + foe 3 = for y, we obtain y — ax ± V '+ 9; show that /(a; + A) - /(a;) = (3a; 2 + 4x)h -f (3a + 2)h 2 + h\ 5. Given f(x) = Vx; prove that f {x + h)-f(x)=--=t-—. Vx -\-h + Vx 6. Given w = — ; prove that Au — r — — =-. x x —j— aj/t- VARIATION. 11. Proportional Variation. One quantity is said to vary proportionally with, or to vary as, another when the ratio of the one to the other remains constant. The sign of variation is a. Thus, y varies as x is written y (xx. Illustrations. 1. The cost per yard of cloth remaining the same, the entire cost (y) varies as the quantity (x). That is, !f«» (1) 2. The space (s) described by a body moving with a uniform velocity (v) varies as the time (t). Or • s oc t (2) 3. The area (u) of a rectangle having a constant altitude (a) varies as the base (x). Or u oc x (3) 12. Principles. I. // one quantity varies as another, one of them is a constant multiple of the other. Let y oc x, then — = m, a constant: x hence, y — mx. FUNDAMENTAL PRINCIPLES. 9 Coe. I. The variations (1), (2) and (3), Art. 11, may be written, respectively, y = mx, where m is the price per yard of cloth ; s = mt, where m is the velocity of the body; u — mx, where m is the altitude of the rectangle. 13. II. If one variable is equal to a constant multiple of another, the first varies as the second. Let y = mx, where m is a constant. Then y £- = m, or y P' = mji + m 2 /* 2 , w 2 /^ 2 = tP', which is a quantity that vanishes with h. DIFFERENTIALS AND ACCELERATIONS. 25. The Differential of a function is that part of its incre- ment which varies proportionally with the increment of the independent variable, and the Acceleration is that part which varies disproportionally with the increment of that variable. Thus, Art. 20, (1), the differential of y is mji, and the accel- eration of y is mji-. The differential of a quantity is denoted by writing the letter d before it. Thus, dy is not d X y, but the differential of y, and is so read. The differentials of functions like x c x* + Ix, and Vl 4- x\ are denoted by d(x a ), d{x 2 -f 7a?) , and d(Vl + a; 2 ). Similarly, the acceleration of a function will often be denoted by writing the letter a before it; as, ay, which is read, "the ac- celeration of y." Cor. I. The increment of a function is equal to the sum of its differential and its acceleration. Thus, dy = dy + ay. 14 DIFFERENTIAL AND INTEGRAL CALCULUS. The acceleration may be positive or negative. Cor. II. When x and y are independent variables, Ax = dx, and Ay = dy. The independent variable may be supposed to change in any manner whatever; its increments are arbitrary, and are themselves independent variables dependent not even on the value of the independent variable itself, while the increments of the depend- ent variable depend on both the independent variable and its increments. This arbitrary character of the independent varia- ble leaves us free to make the most convenient supposition with reference to the manner of its variation, which is that this variation is uniform, or that its increments have no acceleration but are differentials. Cor. III. The differential of a function at any value is what its succeeding increment would be if at that value its chauge became proportional to that of the increment of the independent variable. That is, when y = f(x), (1) the limit of -j-, as h approaches 0, is 1, and (2) dy oc h. For example, in Fig. 4, DP' we have (1) the limit of -jrr, as h approaches 0, is 1, and (2) Dt oc PD. Since the differential of the function varies as the increment of the independent variable, the former will vary uniformly when the latter does. Cor. IV. The differential of a function is positive or negative according as the function is increasing or decreasing. Cor. V. In the increment Ay — mji -\- mjt~, dy = mji = m x dx, and ay — mji\ Hence in Fig. 4, since Ay — DP', and dy — Dt, we have ay = tP'. d u 26. In the equation dy = m x dx, m 1 or -— is called the Derivative or Differential Coefficient of y with respect to x, and is equal to tan XTP, Fig. 4. FUNDAMENTAL PRINCIPLES. 15 Aii 27. Since the limit of -A as Ax approaches 0, = m l9 Art. 24, and since -— ■ = m l3 we have limit r^n dj, _di ■ Jx = 0]_Jx_j dx' pd'*^-*- Aii which is read, " the limit of —^-, as Ax approaches 0, is equal dii to -j-" We are to understand by this that the ratio of the pro- portional variations of y and x can be found by taking the limit of -p. The student should note that the limits of Ay and Ax are not dy and dx; but the limit of -~- is equal to -—, because each is equal to m l3 just as ^f is equal to ^-f, because each is equal to f . 28. In rinding the differential of a function it is not neces- sary to find the entire increment; only the part, or parts, involv- ing the first power of h or dx, will be sufficient, for the terms which involve the higher powers of h form the acceleration of the function. As an example let us find the differential of u = x* - bx* -f 3x (1) Increasing x by h, we have u + Au = (x + ny - 5(x -f ny + 3(« + k) = x* + 4:x*h -f- etc. - b(x z + ?>x 2 h + etc. + 3 (a -f A), (2) where the omitted terms contain higher powers of li. ('2)-(l), tf« = (4a; 3 - 15a; 2 -f 3)dx, Ans. 16 DIFFERENTIAL AND INTEGRAL CALCULUS. 29. Cor. I. If u and v are functions of x, when x is increased by li the proportional increments of u and v are du and dv, and these are the parts of Au and Av which involve the first power of A. DIFFERENTIALS OF GEOMETRIC FUNCTIONS. 30. Differentiation is the operation of finding the differen- tial of a function in terms of the differential of its variable. The process consists in finding the increment of the function and removing from it the acceleration, or in determining what the entire increment would be if it varied as the increment of the variable. The following important formulas are deduced at this time more especially for the purpose of illustrating the preceding principles. 31. Differential of Plane Areas in Rectangular Co-ordinates. Let APP' be any plane curve, OB — x, BP = y, and area of OB PA = u; it is required to find the differential (du) of u. When x is increased by BC (= h), we have b c Fig. 5. Au = BCP'P = BCDP + PDP', which corresponds in form with Ay = mji -f- mjf, Art. 24, in which mji = BCDP, mjtf = PDP',. m x = BP = y, and mjf vanishes with h. Since the initial side of Au is BP (= y), the rectangle BCDP (= yli) is what Au would be had it varied as h; hence the area of the rectangle is the differential of u. '. du = yli or ydx. FUNDAMENTAL PRINCIPLES. 17 Or thus: since the limit of BCP'P -r- BCDP, as BC ap- proaches 0, is 1, and since BCDP oc 2? (7, BCDP (= ?/^) is the differential of 0£P^4 (= %). JfeMotf % Ztmto. The increment BCP'P is > BCDP and the cylinder generated by BCDP (= nyVi) is what the incre- ment Av would be had it varied as h. dv = 7ty*h or 7ry*dx. Method by Limits. Vol. gen'd by BCDP < Av < vol. gen'd by BCP'Q, or ny^Ax < Av < n{y -f- Ay) 2 Ax. Av ^ < ~Ax < ^ + Ay ^' Passing to limits, as in previous examples, we obtain dv = rcy^dx. Hence, the required differential is n times the square of the value of the ordinate expressed in terms of x, multiplied by the differential of the abscissa. Cor. I. In Fig. 5, av, the acceleration, is the volume gener- ated by PDP f . 33. Differential of the Are of a Plane Curve in Rectangu- lar Co-ordinates. In Fig. 4, let s = the length of the arc AP, then PP f = As. Since Dt is what Ay would be had it varied as h, Pt is what As would be had it varied as7^; hence. Dt = dy and Pt — ds; and since Pt — VPD* + Dt% we have ds Vdx' + dif, or (|/l4-J^W Hence, the required differential is the square root of the sum of one and the square of the value of the derivative of y with respect to x, expressed in terms of x, multiplied by the differential ofx. FUNDAMENTAL PRINCIPLES. 19 Cor. I. The limit of the ratio of an arc of any plane curve to its chord is unity. For (Fig. 4), As_ arc PP' _ As _ Ax chord PP' ~ vz? + Zjf " i / 1 + (AyY 9 the limit of which is evidently unity. 34. Differential of Surfaces of Revolution in Rectangular Co-ordinates. Let s = the length of the arc AP, and 8 = the area of the surface generated y by the revolution of AP about OX as an axis, then AS = the area of the surface generated by the revolution of PP' {= As) about BC (= h). At P and P' draw PG and P'Q each equal to PP' and parallel to OX; the areas of the surfaces generated by revolving PG and P'Q about OX are 2nyAs and 27r(y -f- Ay)As, aud evidently ^7$ lies between the former and the latter. That is, Q F yS P A B Fig. 6. 2ityAs < AS < 2n{ij + Ay) As. AS As li approaches 0, the limit of Ay is 0, and the limit of -r- is equal to -=— ; hence we have il Sf 2ity < -=— < 2 Try, or d# = Znyds, or dS = 2ny Vdx 2 + 20 DIFFERENTIAL AND INTEGRAL CALCULUS. Hence, the required differential is the product of the circum- ference of a circle tuhose radius is y, by the differential of the arc of the generating curve. 35. Differential of Plane Areas in Polar Co-ordinates. Let APP' be any plane curve, and let be the pole7 OP (= r) P the radius vector, and . put 6 -- XOP and u = the area of OAP; it is required to find the differential of u. When 6 is increased by the angle POP' (= A 6), u will be a a increased by the area of POP' FlG - 7 - (= Au). From as a centre with the radius Oa (= 1) describe the arc "be (= Ad or d6), and with the radius OP (= r) describe the arc PD (= rdB). Since OP is the initial side of Au, the area of the sector POD is what Au would be had it varied as Ad. du = iOP X PD = ir'dd. POP' Or thus: since the limit of T> nn , as be approaches 0, is 1, and since POD oc be, POD ( = h'dd) is the differential of OAP (= u). Cor. I. The acceleration of u is the area of PDP'. CHAPTER II. ELEMENTARY DIFFERENTIATION AND INTEGRATION. DIFFERENTIATION. 36. Every function may be differentiated by the principles heretofore established, but in practice it is better to use rules, which we now proceed to deduce. 37. To differentiate a constant. Since a constant has no increment, its differential is 0. That is, dc — 0, and dC = 0. Hence, the differential of a constant is 0. 38. To differentiate the product of a constant by a variable. Let u = cv, (1) where c is a constant and v is a function of x. Let h represent any variable increment of x estimated from any particular value of x, and du and dv the corresponding pro- portional increments of u and v, Art. 29; then u + du = c(v -\- dv) = cv -\- cdv (2) (2) - (1), du = cdv (3) .*. Rule. — Multiply the constant by the differential of the variable. 3 9. To differentiate the product of two variables. Let u = vy, (1) where v and y are functions of x. 21 22 DIFFERENTIAL AND INTEGRAL CALCULUS. Let x 9 represent any particular value of x, and u 9 , v 9 , y 9 the corresponding values of u, v, and y; then «' = ^'2/' (i) * Let h represent any variable increment of x estimated from x', and die, dv, and dy the corresponding proportional incre- ments of u, v, and y (Art. 29); then u' + du = {v 9 + dv){y' + dy) - (dv){dy) = Vy + y'efo 4- v'dy (2) The term (dv)(dy) is eliminated, or dropped from the second side, since it varies disproportionally with h (Art. 17) and is therefore a part of the acceleration of u. Subtracting (1) from (2) we have du = y'dv -f- v'dy. Since y' and v 9 are any corresponding values of y and v, we have du — d(vy) = vdy + ydv. .'. Rule. — Multiply the -first by the differential of the second, and the second by the differential of the first, and add the two products. 40- To differentiate the product of any number of variables. Let u — vyz, where v, y, and z are functions of x. Assume w = yz, therefore u = viv\ then du = (w)dv -\- v{dw), also die = zdy -j- ydz; therefore du — (yz)dv -j- v(zdy + ydz) = yzdv -\- vzdy + fl^ffe. In a similar manner it may be shown that d(vtvyz) — vwydz -f- viozdy -f- vyzdw -j- wyzdv. ELEMENTARY DIFFERENTIATION AND INTEGRATION. 23 .*. Eule. — Take the sum of the products obtained by multi- plying the differential of each by all the other variables. 41. To differentiate a traction. Let u = — , where v and y are functions of x, y v Since u — — , we have uxi = v. y Differentiating, udy + ydu = dv. (Art. .39) , 7 dv dy TT ., dv — udy y ° Whence, clu = = - y y __ ydv — vdy ~ ? • .". Rule. — Multiply the denominator by the differential of the numerator and from the product subtract the numerator multiplied by the differential of the denominator, and divide the result by the square of the denominator. 42. Cor. I. The differential of a fraction whose numerator is a constant is minus the numerator into the differential of the denominator divided by the square of the denominator. For, Art. 41, dl- ) = y ° ~ CC y = — ^-, since dc = 0. V y y 43. To differentiate a variable having a constant ex- ponent. Let u = v n , where v is any function of x, and n is any con- stant integer or fraction. I. When the exponent is a positive integer. Since v n = v . v . v . . . to n factors, d(v n ) = v n -\lv + v n ~ l dv ... to ^ terms (Art. 40) = nv n ~ l dv. 24 DIFFERENTIAL AND INTEGRAL CALCULUS. II. When the exponent is a positive fraction. a Let u ~ v c , then u c = v a . Differentiating, cu°~ x du = av a ~ x dv. Substituting for m, cv c ^ = av a ~ x dv. a a- J Reducing, cv da = av a ~ 1 dv. Dividing by «v " " ' , du = jT 1 ** III. When the exponent is negative. Let u — v~ m , then u — — -. v m mv m ] Differentiating, Art. 42, du = ^r^ v Hence, for all the cases, we have this Rule. — Take the product of the exponent, the variable with its exponent diminished by 1, and the differential of the variable. Thus: d(x b ) = bx'dx, d(u~ 7 ) = - 7tT 8 d«, d(v % ) = \vHv y d(%~ 1 ) = - \z~ h dz, 3 V7J' The same rule holds when the exponent n is irrational or imaginary. 44. Cor. I. The differential of the square root of a variable is the differential of the variable divided by twice the radical. ELEMENTARY DIFFERENTIATION AND INTEGRATION 25 For, d{ tfv) = d(tf) = iv^dv = —— 2 Vv' 45. To differentiate the algebraic sum of several variables. Let u = v -\~y — z, (1) where v, y, and z are functions of x. Let h represent any variable increment of x estimated from any particular value of x, and du, dv, dy, and dz the correspond- ing proportional increments of u, v, y, and z; then u + du = v + dv + y + dy — (z -f dz). . . (2) (2) - (1), du = dv + dy- dz. .-. Eule. — Take the algebraic sum of their differentials. EXAMPLES. Differentiate the following: 1. y — x 3 + 5a; 2 - 3x + 7. dy = d(x 3 ) + d(bx*) - d(Sx) + d(7) (Art. 45) = d(x 3 ) + 5d(x*) - Sd(x) + d(l) (Art. 38) = 3x*dx -f lOxdx — 3dx + (Arts. 37, 43) = (3a; 2 + 10a; - 3)dx, Ans. 2. y = x 2 -f 53: + 3. = (4aT s - ar 8 )da;. X 15. «/ 2 = 4aa\ dy — — dx. 16. ?/ 2 + ^ 2 ^i? 2 . rfy= - ? 2 - l)(a: 4 4- 1). (6a; 5 - 4a; 3 4- 2x)dx. 23. dl$x h y k ). 3x~*y h dx 4- 2x h y~ l dy. 24. d[ar a (l 4- a" 3 )]. (- 2ar 3 - 5x «)dx. 25. ^(14-a;)(:c + a; 2 )a;. 26. d(l + 2x*)(l + 4x % ). 27. a> 4- 1) (a; 3 - a; 2 4- x - 1). 4a; 3 o\r. 28. d(x* 4- a)(3x 2 4- 6). (15a- 4 4- 36a- 2 4- 6ax)dx. 29. o'(12a:V 4- Voabe). I8x h i/dx 4- l^ifdy. Differentiate the following: 30. u = \±*> 1 — a; ELEMENT AR T DIFFERENTIA TION AND INTEGRA TION 27 du= a-*w+*)-a+*m-*) (Art41) (1 — x) v ' 2dx , Ans. , ' adx dU = 7T~. -(1-aT 31. a — x X 32. l + x u = —-. — a . 1 -\-x* 33. X* 11 - (i + *r 34. x 3 x 1 U - x* - 1 x-1 35. x 3 ~ a 2 — x 2 ' 36. 2a 2 — 3 (1 - 2a - a 2 )da *» = — (FT**— 7 3 a 2 da du 2xdx du - 8*' + 6s+12 4a + a 2 * tf *~ (4a + a 2 )* ^ 3 (a + to) 3 * a 3 7 3a 2 4- a; 3 _ 0+^- *• = (T+iV* 5, 39. m = fa 2 — 3a + 5. d(a 2 -3a+5) (2a-3Wa /A , JJX dw = — . = \ - J (Art. 44) 2Vx* - 3a + 5 2 fa 2 - 3a + 5 40. w = VW^b. du = — I^L=. V 3a 2 — 5 . . . 2 4-% 41. u = f4a + i 7 9a 3 . dw = V fe 2 fa 2 -f 3a 42. « = afl 4- a. f??t = — — dx. 2 Vl + a 43. w = — - dw Vl - a 2 (1 - x *f A A i/l 4" X 7 ^ 44. ?« •= \ —± — . du = == 1 ~ x {l-x)Vl-x> 28 DIFFERENTIAL AND INTEGRAL CALCULUS. 45. v = (x a -3a? a + 4a;- 5) 5 . du = 5 (a; 3 - 3a; 2 + 4a; - 5)W(a; 3 - 3a; 2 + 4x - 5) (Art. 43) = 5 (a; 3 - 3a; 2 -f 4a; - 5) 4 (3a; a - Gx -f 4)<7a;. 46. w = (a; 3 - 7a; -f 5) 3 . dw = 3(a; 3 -7a;+5) 2 (3a; a -7)da\ 47. u - (x 2 + 5a; - 9) 1 . du = ^(x' i -{-5x-9) i (2x-\-b)dx t 3dx 48. it = (2Vx+ 3)~ 3 . V2iC 3 - 6z + 5> 21. f[- —^~ 3 W VSz 2 - 6* + 5 + G. ELEMENT AR T DIFFERENTIA TION AND INTEGRA TION 35 22. yV + 3z 2 - §)\x 2 + 2x)dx. f (x s + 3:c 2 - 6) § + £ /* xdx 23. / ? . J (1 -z 2 )*. 25 »/ 6rr 2 ^ V^z" — 5' 5+ (7. 4/1-3 £-2 dx~^f [dxj ~~^j r ~ir t = |/l + i^Jdx = l/l + 9 -^dx = tift+toZdx. 20. The differential of the equation of a certain curve is dy = Vx* — 1 dx; find the length of the curve, beginning at the origin. Here | = <&=!; • •• (g)'+l = .«; . = ^ 21. Find the length of the arc of a curve whose equation is y = §(# — 1) § , measured from the point where x = 1. fa; 2 — f. 22. Find the length of a curve the differential of whose equation is dy = \ f x* -f 2xdx, beginning at the origin. fcP + x. x z 1 23. Find the length of the curve y = — -4- ■-, between the limits (1), x = 2, # = 3; (2), x — a, x — h. m i- (») h *~ a * i h ~ a r/s ELEMENTARY DIFFERENTIATION AND INTEGRATION. 41 24. Find the general length of the curve y = f 1 — ^W%, estimated from the origin, / x \ ,- 65. Areas of Surfaces of Revolution. 25. AP is the arc of the circle 2/ 2 = i2 2 —£ 2 ; A find the area of the zone generated by re- volving AP about the axis OX. By Art. 34, dS = %ny\ 1 + l-p) dx; i V: (dy this example, since y* = B? Hence, dy ' dx dS = 27ty]/ 1 + — a dx = 2nRdx. y ■/. .-. S = 2?rE / dx = ZttRx. 26. In ex. 25 find the area of the zone generated by the arc PP' ', supposing OB = a and OE — I. \gf — 27iR(h — a\ ' a 27. In ex. 25, find the surface of the generated sphere. [S] + _l= 4R*\ 28. Find the surface of the cone which is generated by revolving the line y = mx about the axis OX. [Sf= ?ryVx 2 + y\ 29. Find the surface of the paraboloid, the generating curve being the parabola y* = 4ax. [Sf= frVa [(« + *)• -(«)»]. 30. Find the surface generated by the revolution of the curve y = ax 3 about the x axis. x |~(1 + 9a V)' — 11 42 DIFFERENTIA^ AND INTEGRAL CALCULUS. 66. Volumes of Solids of Revolution. 31. Find the volume generated by revolving OBPA about the axis OX, the equation of the curve AP being if = 'Sz 2 - 36z + 105. By Art. 32, civ = ny*dx\ in this example y 2 =3x 2 — 362+ 105, hence Fig. 11. dv = 7t(3x 2 — 362 -f 10b) dx. ,-.v =7i f\sx 2 - 36z -j- im)dx = tt(x* - 18x 2 -f 105.?;). * 32. In the same example find the volume of the solid generated by the revolution (1) of OEA; (2) of EGF. (1) [ v f= 200tt; (2) [v] = - 4tt. 33. Find the volume of the cone which is generated by revolving about x the triangle whose base is x and whose altitude is y (= mx). r x . „ 2X 7 ttw/V Jx\ J v ' v = 7t J (mx)dx = — - — = ny 2 - J. 34. Find the volume of the solid which is generated by revolving y = x* — 4 about OX. v = 7tf(x* - 8a; 2 + 16)^ = 7z[\x" - \x* + 16a?] -j- C. 35. In ex. 34, find the volume generated by the negative area of the given curve. r -i+ 2 __ 34 _ 2 n 3G. Find the volume of the spherical' segments generated (1) by OBPA and (2) by BEP'P, Fig. 10; also find (3) the volume of the sphere (see Ex. 26). (1) [v]=7c(R*a-ia>); (2) |>f= 7t[R\b - a) - W~a*)]; +R (3) [v]_ b = %nR\ 37. Find the volume of a prolate spheroid, the generatrix being the ellipse a 2 y % = a 2 6 2 — £V. M + " = knob 2 CHAPTER III. SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. SUCCESSIVE DIFFERENTIALS. 67. The differential of any function, as u = f(x), denoted by du, is the first differential. The differential of the first differ- ential, viz., d(du), denoted by d 2 u, is the second differential. The differential of the second differential, viz., d(d 2 tt), denoted by d 3 ti, is the third differ ent i al ; and in general the differential of d n ~ l u, denoted by d n u, is the nth differential. du, d 2 ti, d 3 u, etc., are the Successive Differentials of it. 68. If x is the independent variable, its differential (dx) is altogether independent of #,— the increment given x at any in- stant being entirely independent of the value which x may have at that instant: It at once follows, therefore, that the differential of dx with respect to x, like the differential of any other variable which is independent of x, is 0. Hence d(dx) = d' 2 x = 0. For example, take the function u — x 3 ; then (1) du = 3x^X1 (2) d\i = d^x^dx + Zx'didx) = 6xdx'; (3) d 3 u = d\§x)dx* + 6xd(dx") = 6dx 3 ; (4) d*u = d(6)dx 3 + 6d{dx 3 ) = 0. Therefore, in finding the successive differentials of a func- tion, we treat the differential of the independent variable as a constant. See Appendix, A 3 . The student should observe the difference between expres- sions like d 2 y, dy*, and d(y' i ); d 2 y is the second differential of y, dy' 2 is the square of the differential of y, and d(y*) is the differ- ential of y 2 , or 2ydy, 43 44 DIFFERENTIAL AND INTEGRAL CALCULUS, If u = f(x), the successive derivatives or differential coef- jicients ox w with respect to sc are -7-, -r-^, y-j, etc. The successive derivatives of /(#) are also denoted by f'(x), f"(x),f"(x),...r(x). Therefore, when u = f(x), we have £=/», £=/», ••• S -/-(*)• EXAMPLES. 1. Find the successive differentials ofy = x\ dy = 4x*dx. d*y = d(±x*dx) = 4tdxd(x % ) = \2x*dx*. d z y = d{12x i dx i ) = 12da?d(x*) ^ 2ixdx\ d'y — d(24xdx*) = 24dx 3 d(x) — 24dx\ d b y = d(2±dx A ) = 0. 2. Find the successive derivatives of x % — 4x 2 -j- 3x — 5. Let /"(#) = a; 3 — 4a; 2 -f 3a; — 5 ; then /'(a?) = j (x* - 4a; a + 3x - 5) = 3a; a - 8a; + 3; f'(x) = ~ (3a; 2 - 8x + 3) = 6x - 8; /"(*)=J;( — 7= in. per second. 1/22 7. A circular plate of metal expands by heat so that its dia- meter increases uniformly at the rate of 2 inches per second; at what rate is the surface increasing when the diameter is 5 inches ? Let x = the diameter, u = the area; then u = -x*. bn sq. in. per second. 8. In the last problem, if the surface increases uniformly at the rate of 50 sq. inches per second, at what rate will the diam- eter be increasing when it becomes 5 inches ? 20 . — m. per second. 7t r 9. The volume of a spherical soap-bubble increases how many times as fast as the radius ? When its radius is 4 in., and increasing at the rate of i in. per second, how fast is the volume increasing ? Let x = the radius, u — volume; then u = ^7tx 3 . (1) 4:7tx 2 times as fast. (2) 32^r cu. in. per second. 10. A ladder 50 ft. long is leaning against a perpendicular wall, the foot of the ladder being on a horizontal plane x ft. from the base of the wall. Suppose the foot of the ladder to be pulled away from the wall at the rate of 3 ft. per minute. (1) How fast is the top of the ladder descending when x = 14 ft.? (2) How fast is it descending when x = 30 ft ? (3) What is the value of x when the top of the ladder is descending at the rate SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. 49 of 4 ft. per minute ? (4) And what at the time the bottom and top of the ladder are moving at the same rate ? Let y — the distance from the base of the wall to the top of the ladder; then y = V2500 — x>- (1) J ft. per minute. (2) 2% ft. per minute. (3) x = 40 ft. (4) x = 25 V% ft. 11. What is the value of x at the point where x 3 — 5z 2 -j- 17:t and x 3 — 3.r change at the same rate ? x —2. 12. Find the values of x at the points where the rate of change of x* — 12^ 2 -f- 45.T — 13 is zero. x = 3, x = 5. 13. In a parabola whose equation is y 1 = 12a', if x increases uniformly at the rate of 2 in. per second, at what rate is y in- creasing when x = 3 inches ? 2 in. per second. 14. In the same parabola, at what point do y and x vary at the same rate ? When y = 6. 15. In the ellipse whose equation is 44f# 2 -{- 25# 2 = 1111|, at what point of the curve does y decrease at the same rate that x increases ? When y = 3 and x — 5^. 16. Find the points where the rate of change of the ordinate y = x* — Gar + ox -j- 5 is equal to the rate of change of the slope of the curve. Where x = 1 and x = 5. 17. Two straight roads intersect at right angles; a bicyclist travelling the one at the rate of 10 miles per hour passes the intersection 2^- hours in advance of another travelling the other road at the rate of 8 miles per hour. At what rate were they separating (1) at the end of H hours after the first man passed the intersection? (2) At the end of 2-J- hours? (3) Eeqnired the distance (y) between them when it is not changing. (i) % = 5tV; (2) % = 10; (?) y = W fS mi. 72. Applications to Geometry. The rates of change of the areas and lengths of carves and of the areas and volumes of surfaces and solids of revolution are given by Arts. 31, 32, 33, 34. The applications are made as in Arts. 62, 64, 65, 6Q. 50 DIFFERENTIAL AND INTEGRAL CALCULUS. Rate of Change of Curves- Formula, ds = y 1 -f \-f-\ dx. 18. Find the rate of change of the arc of the parabola y* = lax. ds — y 1 -j — dx. 19. In the previous example, if a = 9, and x increases at the rate of 12 inches per second, at what rate will the arc be increas- ing when x = 16 inches ? ds = 15 in. per second. 20. Show that the rate of change of the arc of the circle I I , DS '■• 7 R(lX x^ + if— IF is ds = - 21. In a circle whose radius is 20 in., the abscissa changes at the rate of n in. per sec; at what rate is the arc increasing when x = 12 in. ? \n in. per second. 73. Given the rate of change of the arc of a curve (ds), to find the rates of change of its co-ordinates x and y. Take the parabola y 2 = lax, then ydy = 2adx; between this equation and ds' 2 = dx* -\- dy 1 eliminate (1) dx and (2) dy, and we have dy = — — - ds and dx — y ds. Vla^-t-tf r a-x-x 22. In the parabola y' ==■ Ix, if the arc increases uniformly at the rate of 5 inches per second, at what rates are y and x in- creasing when x — 9 inches ? i |/10 and | ^10 inches per second. 23. If the arc of the circle x 1 -\- y 1 = 100 increases at the rate of 5 inches per second, at what rates are y and x changing when x = G inches ? — 3 and -f- 4 in. per second. APPLICATION TO MECHANICS. 74. Velocity is the rate of change of the distance described by a moving body. Hence, if s = the distance, v == the velocity and t — the time, we have I. v = -rrl .'. s — I vdt. and t — I — . dt «/ J v SUCCESSIVE DIFFERENTIALS AND BATE OF CHANGE. 51 Again, denoting the rate of change of the velocity by a', we have II. a' = ^r dv d 2 dt dt ■i > ,-. v = I a'dt, and t — I — . EXAMPLES. 1. If 6* = 3f, what is the velocity and its rate of change ? ds Since s — 3f, —- — Qt = v, the velocity; dv and since v = 6t, —rr = 6 = #', the rate of change of v. Clo Thus, if the unit of s is one foot, and the unit of t one ds dv second, v = — [= Qt] ft. per second, and a f = -jy [=6] ft. per clt civ second. 2. A body passes over a distance of ct* in t seconds; find v and a', (1) in general, and (2) at the end of 9 seconds. (1) — *— and C —\ (2) — and - — . 2 Vt 4 Vf W 6 108 3. A body after moving t seconds has a velocity of 3f -f- 2^ ft. per second; find its distance from the point of starting. s =J*vdt = ?+ t\ 4. The velocity of a body after moving t seconds is 5f ft. per second; (1) how far will it be from the point of starting in 3 seconds ? (2) In what time will it pass over a distance of 360 feet? (1) 45 ft.; (2) 6 sec. 5. A body moves from A, and in t seconds its velocity is 14/^ ft. per second; (1) how far is the body from A ? (2) In how many seconds will the body have gone 847 feet ? (1) If ft.; (2) 11 seconds. 75. The velocity is positive or negative according as s is in- creasing or decreasing, and a' sustains the same relation to v; 52 DIFFERENTIAL AND INTEGRAL CALCULUS. therefore, if s increases as the moving' body goes forward and decreases as it goes backivard, the body is moving forward or backward according as v isiwsitive or negative. 6. A train left a station and in * hours was at a distance of \tf — 4* 3 4- 16^ 2 miles from the starting-point; required the velocity and its rate of change, also when the train was moving backward, when the velocity or rate per hour was decreasing, and the entire distance travelled in 10 hours. s = it* — 4* 3 -j- 16* 2 — the distance from station. j = f - 12* 2 + 32* = v = the velocity. dv — = 3*' - 24* + 32 = a' = the rate of change of v. Civ The roots of * 3 — 12*' + 32* = are 0, 4 and 8; therefore v is negative, and the train was moving backward from the 4th to the 8th hour. Again, the roots of 3f - 24* + 32 = are 1.7 - and 6.3 -f ; hence a' is negative, and therefore v was decreasing, from the 1.7th to the 6.3th hour. The roots of It* - 4f + 16* 2 = are 0, 0, 8 and 8; that is, s = when * = 8; hence the train was at the starting-point at the end of 8 hours, having gone backward as far as it had forward. Since the train moved forward the first four hours, then backward the next four hours, and then forward, the entire dis- tance passed over in 10 hours was M - M + M = 64 + 64 + 100 = 228 (miles). 7. A train left a station and in * hours was moving at the rate of t 3 — 21f -j- 80* miles per hour; required (1) the distance from the starting-point; (2) when the train was moving back- ward; (3) when its rate per hour was decreasing; (4) when the train repassed the station; and (5) how far it had travelled when it passed the starting-point the last time. SUCCESSIVE DIFFERENTIALS A ND RATE OF CHANGE. 53 (1) s =fvdt =yV - 21*' + 80t)dt = it* - 7f -f ±0t* (2) From the 5th to the 16th hour; (3) from the 2.27th to the 11.72th hour; (4) in 8 and 20 hours; 5 16 20 (5) M - \s] + [si = 4658^ miles. 5 16 8. A traveller left a point A at 12 m., and in t hours after his rate per hour was 5 — t miles; (1) how far forward did he go? (2) At what times was he 8 miles from A ? (3) What were his rates per hour when at a distance of 10J miles from A ? (1) 12| miles; (2) 2 p.m. and 8 p.m.; (3) +2 mi. and - 2 mi. per hour. 76. Uniformly Accelerated Motion is that in which the rate of change of the velocity («') is constant. That is, v changes Av dv uniformly or v oc t; hence, Art. 70, — - = -— = a' = the rate of change of the velocity. Formulas. v =Ja'dt = a't + C = a't + v , . . (1) and s =J vdt =J (a't + v )dt=ia't* +v t + s , . (2) in which v and s represent the initial velocity and distance; that is, the values of v and s when t = 0. If v — and s = when t = 0, then (1) and (2) become /2s — and v — V2a's. . (3) 77. The increment of v, or acceleration, produced by gravity is about 32.17 ft. per second, and is usually represented by g. Hence, by substituting g for a' in (3), we obtain the four for- mulas for the free fall of bodies in vacuo near the earth's sur- face. When the bodies are not in vacuo the formulas generally are slightly inaccurate, on account of the resistance of the at- mosphere. 54 DIFFERENTIAL AND INTEGRAL CALCULUS. PROBLEMS. 1. A rifle-ball is projected from in the direction of Fwith a velocity of b ft. per second; required its path, knowing that its velocity in t seconds along the action-line of gravity (OX) will be gt ft. per second. Let OX and OF be the axis of x and y, respectively; then -~ = b, and — = gt; dt x, y — M, . . . (1) and x = \gt\ . . (2) Eliminating t between (1) and (2), we have Fig. 13. J g that is, the path of the ball is an arc of a parabola. 2. A body starts from 0, and in t seconds its velocity in the direction of OX is 2abt, and in the direction of OY is a?f — b*; find its velocity along its path Onm, the distances in the direction of each axis and along the line of its path, and the equation of its path, the axes being rectangular. Let v X9 v y and v s represent respectively the velocities in the directions of the axes x and y and the path s. Then >. x = J2abtdt = abP; . \ y =yV^ - ^) (U = 4* s=f{aH 2 +byit = lan % -\-bn. . . . . Vt\ aH* (1) (2) b\ (3) Now to find the path of the body, we eliminate t between (1) and (2) and obtain CHAPTER IV. GENERAL DIFFERENTIATION. LOGARITHMS. 78. Lemma. The limit of \l-\ — ) ,as z approaches infinity, 1_ . 1 . 1 is the sum of the infinite series 2+-jx- + -r^+T7 + ^ c Assuming the binomial theorem for positive integral values of z, we have -+'+W-!-)+i( i -J)( i -:-)+"- which evidently approaches 2 + -— + — - -f etc., as z approaches If. I- infinity. The sum of this series is represented by e. It is the base of the natural system of logarithms, which is equal to 2.718281, approximately. mit (i+iy=, - CO \ % ) limit z — 79. To differentiate the logarithm of a variable. Let u = log a v, where v is a function of x. 55 56 DIFFERENTIAL AND INTEGRAL CALCULUS. When x is increased by h we have An — log a (v + Av) — log a v (v + ^ V = W — J Au ~Av i = v M 1 + tJ Passing to the limit, remembering that as h approaches 0, — approaches 0, and —r- approaches go , we have, as in the pre- ceding lemma, du 1 , ' .dv * ^7 = v lo Sa e> °r du = (log a e)~-. Rule. — Divide the differential of the variable by the variable itself, and multiply the quotient by the constant log a e. The factor log a e, usually represented by m, is called the modulus of the system whose base is a. When a — e the mod- dv ulus is unity and we have du = — , simply. Herein lies the advantage of the natural system of logarithms, whose base is e, in all discussions of a theoretical nature. Hereafter when the base of a given logarithm is not indicated, it is to be understood that the base is e. EXAMPLES. 1. Differentiate u = log(x* — 2x + 5). _ d(x* - 2x + 5) __ (3 % a - 2)dx x 3 - 2x + 5 ~ x 3 - 2x + 5* * For another method, see Appendix, A\ , Cor. III. GENERAL DIFFERENTIATION. 57 VI — xl 1 — x 2dx 1 —x 2dx X (1 -xf 1+x \~x % 3x 2 dx 3. u = log t^ 3 - a\ du =. _ 4. m = log a (5z 2 — x 3 )\ du — fta v ' bx — x 5. w = log (x + ^1 + a 2 ). ^ = 6. w = log [(& — x) \/a + »]• ^ = _ 4m (10 — 3x)dx dx (a + 3#)^# 7. y = \og*x. dy — 3log 2 #— . 8. y = log 4 (log x). dy x 4 log 3 (logx)dx x log X 9. y = log (log a:). rf y ^ =__-.. dx 10. 2/ == log (x-\-a-\- i/2ax -\- x*). dy \ f 2ax -j~ # a EXPONENTIAL FUNCTIONS. 80. To differentiate u = r? r . Passing to logarithms, we have log u — v log a. Differentiating, — = dv log #. Multiplying by u, du = u log a dv, Substituting a v for u, du = a v \oga dv* * For another method, see Appendix, At. 58 DIFFERENTIAL AND INTEGRAL CALCULUS. Hence, the differential of an exponential function with a constant base is equal to the function itself into the natural logarithm of the base into the differential of the exponent. Cor. I. d(e v ) = e v dv, since log e = 1. EXAMPLES. 1. Differentiate y = m VI +^\ - dy = m ^log (m)d( V l+^) = ™ V ^ }?*&)*& p \Og Xrfrr 2.y = e l °e*. dy = - — . x 3. y — e x log x . dy — e x log *(log x + l)dx. e x 7 e x x dx 4- y = — • ^ = (jq^-r 5. y — e x (l - x'). dy = e x (l — Zx* — x')dx. l+x e x (l - x 3 ). lOSf . 6 1 + 6 X e* — e~ x , dx 6. y = log — — -. = — Vl — ^ a rfv. tfa; = -^ , or f?(cos -1 v) = —'!— =, Vi - if Vi - y which always has the sign of — sin v. * To avoid the double sign ±, we shall suppose < v < — ; for any other quadrant the sign will be that of cos v. 64 DIFFERENTIAL AND INTEGRAL CALCULUS. 92. To differentiate tan -1 y. Let v = tan -1 y, then y = tan v. ^ = sec 2 v dv — (1 -j- ?/ 2 )^. dv = ^y, or d(tnn- 1 y)= ^ , - In a similar manner we find 93. ^(cofc- 1 : ?y) = 94. ^(sec 1 y) = 95. rf(cosec _1 2/) = 95. ^(vers* 1 ?/) = dy y Vtf - 1 dy V2y - y 9 EXAMPLES. Differentiate the following : \. y — sin -1 Vx. dx d Vx 2 Vx dx dy~ Vl~(VxY Vl-x 2Vx-a? ,., = tan-.(^). J \ - x \ 2dx d _ Vl + J & + '*)" <** "*" Vi-hJ +av (1 + z) 2 3. ?/ = sec -1 wz. «?/ = — x yn*a? — . /CX -j tlJu 4. it — ver -1 — — . dy — ■ J 9 * V$x-a? Vl - X* dx GENERAL DIFFERENTIATION. Q5 5. y = sin -1 (3a; — 4# s ). dy = 6. y = sin -1 (2a; — 1). dy = yx — x 2 ' ¥. y = sin -1 (sin x). dy = dx. 8. y = sin -1 (Vsin a;). *??/ = |(Vl -f- cosec a;)^a;. n , _j 2a; , 2 |/ c « _ ^ _ 2^ _ fl v \ -o J 20. r Wfl gg _ logte+6+Vc u +(^+aa;)'+a 21. / a — —T5-T — j = m tan M + C. ^ cfx 4- 2rtto + ft 2 -j- c 2 \ c I 66 DIFFERENTIAL AND INTEGRAL CALCULUS. r m{ad-bc)dx _ J ax'+b \ ' J (ax + b){cx + tf) ~~ g \cx -\-d)^ The last four equations may be conveniently employed in integrating a certain important class of differentials, of which the following are illustrations: 23. Eequired the integral of x* - 3x - 28 Here x* — 3x — 28 = (x — 7)(x -\- 4); hence we may inte- grate by formula 22, thus : Make ax -{-b = x— 7, ex -f d = x -{- 4, and m(ad — be) = 5; we then have a = 1, b = — 1, c = 1, d = & and m = T 5 T ; hence, substituting in 22, we have /hdx 5 , (x — 7\ , ~ a --to-28 = n log U+lj + a Zdx 24. Eequired the integral of 4z 2 + 3a? + 1 Integrating by Ex. 21, we have a 2 = 4, 2ab = 3, b* + £ a = /> and m#c = 3 ; whence a = 2, J = f, c = i V'7 and wi = — — , V7 f Zdx 6 f -i f 2x + t \ I •* •V4x 2 + 3a; + l~W IT^"/" 1 " 25. Find the integral of Vl6 - 12a? - 4a; 2 Integrating by Ex. 19, we have a? — 4, 2ab = 12, c 5 — 5 2 = 16 and ww = 5 ; hence a = 2, & = 3, c = 5, wi = |. /» 5^ 5 . . /2a? + 3 \ , ' ~ .\ / — r: — ■ = - Sill ^ + C. J Vl6 - 12a? - 4a? 2 2 V o / /• dx 2 , _ x 2a? + 1 C dx 1 .8!^ , - GENERAL DIFFERENTIATION. 67 J 2x l - 4a; - 7 ■■/■ V2 . 2a; 2 - 3 V2 2 + 3 Vi + G 30 32. f dx dx . , 3 + 2x „ sin" 1 ,_ + C. 4/13 Vm 4- ?&r + rx* — log (z Vr + rfz Vm + nx — rx* dx' Vr n 2 Vr sin' Vm + nx + rx'J ■■>- - \rx \/±mr + w + c. m + nx + r# 3 ' V4wir — w a tan" / 2rx \V\.mr — n 2 0. 97. To find the differential of an arc in polar co-ordinates. Let AP (= s) be the arc of a curve, the pole, OP {— r) the radius vector, and PT a tangent to the curve at P. Let 6 = XOP and ip = OPT. Increase by POP' (=40), then arc PP' = Js and OP' = r 4- Jr. Draw PD perpendicular to OP', then PD = r sin J0 and PP' = r 4- 4r — r cos z/#. The chord PP' = VPD 2 + DP", chord PP' _ a/(PD\\ ( dp '\\ Aa f \ AH I 1 Ad I Ad ~ f \Ad J ' \ Ad I Passing to the limit, remembering that as A 6 approaches 0, ,, .. .. . chord PP' ,&mA0 .. DoN . A . the limits of — - — p^- and — ^— (Art. 33), also of cos Ad, is arc Jr ± each unity, we have ds =. J0 r 2 + rfr tf0 1 ^0. 68 DIFFERENTIAL AND INTEGRAL CALCULUS. 98. Cok. I. As P' approaches P, the angle OP'P approaches the angle OPT (=*/>); therefore A , limit \~ PD~] rdd , rdd , dr FUNCTIONS OF TWO OR MORE INDEPENDENT VARIABLES. ^°. A function of two variables, as u = x 2 y -f- y % and u = sin (x + ?/)> is represented by f(x, y); and similarly f(x,y, z) represents a function of the three variables x, y and z. Since x and y are independent of each other, the function u =f(x, y) may change in three ways. Thus, let u = xy = area of <9£ZM, where 0£ = x and 0^4 = ?/. (1) x may change and y not, which would give du = BCDP = ydx; (2) y may change and x not, which would give du = APFG = acty; (3) both # and ?/ may change, d# dec FlG ig which would give (understanding by du the portion of the increment which is of the first degree in dx and dy) du — ydx -\- xdy. Hence du may have three different values, and it is desirable to employ a notation by which they may be represented. 100. A Partial Differential of a function of two or more variables is the differential obtained on the hypothesis that only one of its variables changes, as ydx and xdy in the previous example, and these are denoted respectively by y dx and -j—dy, or d x u and d y u. 101. Total Differential. In a function of two or more in- dependent variables, if each variable receives an increment, that portion of the corresponding increment of the function which is of the first degree with respect to the increments of the variables is the total differential of the function. GENERAL DIFFERENTIATION. 69 102. Prop. The total differential of a function of two or more independent variables is the sum of its partial differentials. Let u represent any function of x and y. When x becomes x -\- h, the part of the corresponding incie- dn ment of u which involves the first power of h or dx is j—dx. ax Hence, omitting the terms involving the higher powers of dx, Art. 29, the new value of u is , du .. u -\- -r-dx. dx In this new value of u when y is increased by h (= dy) the du parts of the corresponding increments of u and —dx which in- ax volve only the first power of dy are -j—dy and -z-[ — dx\dy; hence, omitting the terms involving the higher powers of dy, the second new value of u is , du -. , du , , d (du 7 \ ., u+ dx dx+ a y d y + Ty\ix dx ) (1 y> which result is the same as if x and y had changed simulta- neously, for the result of increasing x by dx and y by dy is evi- dently the same whether the changes be made separately or simultaneously. Hence, since tHe last term involves the product of dx and dy, we have 7 du 7 . du , du= ix dx+ ^ d y- In a similar manner it may be shown that the theorem is true of functions having any number of variables. Cor. I. The total differential of a function is the sum of those parts of its increment which vary as the increments of the variables, respectively. Cok. II. The theorem is also true of functions whose variables are not independent of each other. 70 DIFFERENTIAL AND INTEGRAL CALCULUS. Cor. III. The total differential of a function of two or more variables, as x, y and z, is the sum of the differentials obtained by first differentiating as though x only were variable, and then as though y only were variable, and then as though z only were variable. 103. A Partial Derivative of a function of two or more variables is the ratio of the partial differential of the function to the differential of the variable supposed to change. 104. The Total Derivative of a function of two or more variables, only one of which is independent, is the ratio of the total differential of the function to the differential of the inde- pendent variable. To prevent confusion w r e shall distinguish the total differen- tial or derivative of two or more variables by enclosing it in brackets. Thus, if u =f(x, y), [du] = C ^dx + ^-dy, du du where j—dx and -r-dy are the partial differentials with respect (XX (XiJ to x and y, respectively. EXAMPLES. Find the total differential of — 1. u . = z 2 — Zxy -f 2y*. j—dx — (2x — 3y)dx; -r-dy = — (3x — iy)dy. .-. [du] = {2x - 3y)dx - (3z - ±y)dy. 2. * = *+*. [du] = 2 _M^lJ^i. x-y (x — y) . ,x r 7 .. ydx — xdy 3. u — sin" 1 -. [du] = J1 — / -. V y S/if - x 2 y xdy — ydx GENERAL DIFFERENTIATION. 71 5. u — log y x . 6. u = y Binx . [du] = —dy-\- log ydx. [du] = y sin x log y cos x dx -f- sma; f 105. Function of Functions. If u — F(y) and y — f{x), u is indirectly a function of x through y. In such cases the value du of — may be obtained by finding the value of u in terms of x, and differentiating the result; but it is often more easily found by the formula du _ du dy dx ~ dy dx (i) That is, the derivative of u with respect to x is equal to the derivative of u with respect to y multiplied by the derivative of y with respect to x. Thus, if u = tan -1 y, and y = log x, then du dy i+y' dx and du dx s(i + y % Y 6?# If w — i^(v, y), v =f(x) and ?/ =/,(#), to obtain total derivative of w with respect to x, we may proceed thus du , , the Since dividing by c?a?, dx dv dy *' du dv du dy dv dx dy dx' (2) du . which gives y 1 - in terms of derivatives which can be reckoned cix out from the given equations. Thus, if u = v* -\- vy, v — log x and y = e x , then du _ du dv 1 , — = 2v -f- y, —- — v, -T- = — and ^ J dy dx x dy dx = e x ; 72 DIFFERENTIAL AND INTEGRAL CALCULUS. substituting in (2), we have du 2v + y ~r = L -^- + ve x . ax x If u = F(x, v, z), v —f{x) and z =f l (x), we have r 7 -, du _ du , , du , rWwl _ e?w f?w dv du dz \_dx J ~~ dx dv dx dz dx 9 ^ ' are partial derivatives, and — the , du du .. du where -7—, -r- and -7- cte ow dz total derivative of w. EXAMPLES. Find -j— in the following: 1. u — e x (y — z), y = sin x, and z = cos x. 2. « = tan"' |, and y = ^^^i 5 ! [g] = - -^ 3. u = tan * (#y), and ?/ = e x . Vr 2 - a; 2 * dxj l + zV*' 4. If # = uz and w = e x , z — x* — 4z 3 -f 12z 2 — 24a; -f 24, find the slope of the curve of which y is the ordinate and x the abscissa. dy „ . — --u — e x x . dx 106. Successive Partial Differentials and Derivatives. We have seen that the differential of u (= f{x, y) ) (1) with re- spect to x is denoted by -r-(u)dx, and (2) with respect to y by GENERAL DIFFERENTIATION. 73 Similarty, the differential of —dx d [ du , \ , d 2 u dx* (1) with respect to x is denoted by "i—l—r-dx \dx, , rt v " -. , . t , t , d ( du .. \ _ rZ 2 « rfo *?■?/ (2) with respect to ?/ is denoted by —\-j—ax\dy ~ dx\dx J ' dx~ d ( du \ _ d 2 u d dy \dx ' I J ' dx dy d*n du dx Hence — - -^ — is a symbol for the result obtained by dif- a y ax ferentiating u two times in succession : once, and first, with respect to y, and then once with respect to x. Similarly, - — -z—odx dip indicates the result of three successive differentials dx dif J of u\ first, once with respect to x, and then twice with respect to y. In finding these successive partial differentials, we treat dy and dx as constants, since y and x are regarded as independent variables, see Art. 68. The symbols for the partial derivatives are y + xf; verify gjfr = J^-. 3. If u = y log (1 -f xy), show that dy dx dx dy „. x* — y 2 .. ^ 2 w d 2 u 4. G-iven w = 2 , , - : verify - — —- = - — - . # + V dx dy dy dx 108. To find the successive differentials of a function of two independent variables. Let u = f{x, y); then &$=%**+%** a) dit du Differentiating (1) and observing that j— and -j~ are usually functions of x and y, and that x and y are independent, Art. 68, we get r rfV J = w dx * + Sk dx dy + wk dy dx + w df ' or, Art. 107, Differentiating (2), remembering that each term is a function of x and y, we have [VTwl — ~--dx* 4- 3 7 „ ., dx* dy + 3^ — ^-jdx dy 1 + -r~rdy 9 . . . ; L J ^ 3 ' efo 2 d2/ J ' dxdif * ' d*/ 3 * and similarly may [d\i\, [d*u], etc., be found. By observing the analogy between the values of [d 2 u] and [d*u], and the de- velopments of (a -f- #) 2 and (a -f- ^) 3 , the formula for the value of [rf"«] may be easily written out. 109. Implicit Functions. In functions of the form dii> dit f(x, y) — 0, the formula [du] = j-dx + -r— dy is often useful in dv finding the value of the derivative, or slope, -ys GENERAL DIFFERENTIATION. 75 Thus, take f(x, y) = ax* -f- x sin y = 0. Making u = ax 3 -j- x sm y> we nave \du~\ = -j—dx -f- -7- dy = (Sax 2 -j- sin y)dx -{- x cos y dy. But since u = 0, [du] = 0. du dy _ dx 3ax* -f- sin y dx ~~ du ~ x cos y dy EXAMPLES. du Find the derivative, ~, of the following: -. / zn2 3 . 2 dy 3X* — 2ax 2. *• + 3d*y - «y» = 0. J.' = f + 4 f * . 3.^+3^+^=0, fe = - *;+ < y . 1 ^ ' y dx y* -\- ax 110. Successive Derivatives of an Implicit Function. The following examples will serve to illustrate how the succes- sive derivatives of implicit functions in general may be deter- mined. EXAMPLES. d*y 1. Find ~~ Q when y 2 — kax = 0. ax XT dy 2a Here j = — (1) ax y v ' Differentiating (1), we have d*y _ - 2ady ( dx ~ y* w 76 DIFFERENTIAL AND INTEGRAL CALCULUS. Eliminating dy in (1) and (2), we have ^l - _ 1?! dx 2 ~ if 2. Given a'lf + ftV - « 2 6 2 = 0, show that %%r «= - -jL 3. Given y s + z s - 3ff»y = 0, show that ^f= - -^BL. 111. Change of the Independent Variable. After ob- taining the derivatives ~~, ^~, ~, etc., on the hypothesis that x was the independent variable and y the function, it is some- times desirable to change these expressions into their equivalents with y for the independent variable and x the function, or with x and y for the functions and some other variable, as t, for the independent variable, and so on. dti 112. To find the successive derivatives of ~ when ax neither x nor y is independent. Under this hypothesis -j- is to be differentiated as a fraction having both terms variable. d?y _ d (dy\ _ dx d*y — dyd*x ( . dx* dx\dxj dx 3 cr -i i d 3 y d (cFy Similarly, ^ = &(^ _ (*fo ^? 3 ?/ — ^?/ d 3 x)dx — 3(dx d 2 y — dy d 2 x)d" i x . ~ : ! dx 1 ! ' ' * W In like manner we obtain the other successive derivatives. Cor. I. If y is independent, then d 2 y = d 3 y = 0, and we have d*y dy d*x dx 2 ~ dx 3 ' ' ' ' ' d*y _ 3(d 2 xydy - d*x dy dx dx 3 ~ " dx b (3) (4) GENERAL DIFFERENTIATION. 11 Formulas (1) and (2) give us the values to be substituted for -t4 and y^ when neither x nor y is independent; and formulas (3) and (4) give us the values of the same derivatives when y is independent. If a new variable t, of which x =f(t), is to be the inde- pendent variable, in Art. Ill, we replace x, dx, d 2 x, etc., by their values as determined from x =f(t). EXAMPLES. 1. Given yd 2 y -f- dy 2 + dx 2 == 0, where x is independent, to find (1) the transformed equation in which neither x nor y is independent; also (2) the one in which y is independent. d 2 v (1) Dividing by dx*, substituting for -~ f rom 0)> an d multi- plying both members by dx 3 , we^have y(d 2 y dx — d 2 x dy) + dy 2 dx -\- dx 3 = 0. (2) Making d 2 y = in this last equation, and dividing by — dy 3 , we have d 2 x dx s dx _ y dif dif dy ~ 2. Change the independent variable from x to t in ** + \*!L + y = o, when x = %Vl dx xdx d 2 y Substituting for -~ from (1), multiplying by x dx 3 , and mak- ing x = 2VI, dx — t~ i dt, and d 2 x = — \t~^df, we obtain d 2 y dy Change the independent variable from x to y in the two fol- lowing equations: Wary c?» efc 3 dx 2 \dx) ' dy 3 dy 2 &y dif 3 d\f _ d 2 z dfz 2 _ _ 4 * *^ 2 + dx~ 3 ~ tic ~ °' X df + dy 2 l ~ U * 78 DIFFERENTIAL AND INTEGRAL CALCULUS. Change the independent variable from x to t in the two fol- lowing equations : K d 2 y , 2x dy , y _ , 5 - a£ + r+p I* + (i+iy = °' where • = tan *• 6 - < x - ^S - 41 = °' where * = cos *• S= °- 7. Find the value of R = f 1 + T^j ■*■ ~r^, where x is in- dependent, supposing neither x nor y to be independent. R = J&f+Jgf^ dr ^ 2 «/ — dy *f# MISCELLANEOUS EXAMPLES. Differentiate the following: 1. y = 3(as 2 + l) § (4as 2 - 3). dy = 56x\x 2 + l)*tfas. 1 dy x 2. y = ■ -=. -f- = . - 1. x H- Vl + a; 2 «& Vl + a 1 _ Va + x + ^a — a; dy a*-\-aVd *—x* 4. y = x-\- log cos f — — as). 7r \ dy 9 dx 1 -f- tan a; Va + |/ie dy_ Va 5 ' y - l0g VTZ~^ dx~ {a - x) vV x log X , _ #i . dy lo & ^ „ , A M - x + 1 dy__ g - 1 7. y - log y jq- x + r dx - ^ + x , + r 8. y = sin (as + a) cos (x — a). dy = cos 2a;das. 9. y = log tan (^ + - J. ^ = sec as. 10. sin 2a: = 2 sin as cos x. cos 2as = cos 3 x — sin 2 x. GENERAL DIFFERENTIATION. 79 . n 2 tan x 1 — tan 2 x 11. sm 2z = .,,,„ . cos 2x — — — - — j— . 1 -f- tan a; 1 -j- tan x 12. sin 3a; = 3 sin a; — 4 sin 3 a?. cos 3x = 4 cOs 3 a; — 3 cos a;. ■1 /,CC % 1 13. y = tan ^ aa; e* + e - * ^-lX dy 6x* 14. ?/ = COS M 6 , -, . -T~— « T-T. ^ Va 6 -f 17 dx x e + 1 . 1 dy 2 lO. 1/ = SCO -1 rr-z r. ~-= . J 2a; 2 — 1 dx tfl — a. 16. y = tan" 1 a' + tan" 1 ^-^. ^- = 0. . 1 -j- a; da; Prove the following by differentiation: 17. / — = r — a— ^^ — = tan -1 in tan a;) + C. J cos a) -j- n sin a: v y 1 f*2ax*dx x 1 _ /as — a\ is - y f^t« = tan « +2 lo ^ &+i/ + « 19. lo g (l + z)==*-f+|-f + etc. 20. tan -1 x = x — '-- + — — — -f- etc. o 5 7 Find the slopes of the following curves: 21. The quadratrix, y = (a — x) tan — . dy ?r / . 9 nx , nx -/- = ■— (a — x) sec 2 tan — -. dx 2a 2a 2a y 22. The c} T cloid, x = r vers -1 \/2ry — y 1 dy_ _ J2r - y 80 DIFFERENTIAL AND INTEGRAL CALCULUS. c i - —~\ 23. The catenary, y = -\6 >c -f e 7. (V 24. The tractrix, x = a log -M I- d\j If* -*\ y ?j 2 £- - f fl a - y\ dy _ y 6 x 2 dx y a * _ y * Find the following: ^ 3(x 2 + 2) 3 27. /*(§«* - 5^ § + 2af* + ix~ % )dx. (X ~ 1)3 + G /"iV- ^ ( a » _ x *y 17 z 4 3aV ' 29. f—— 3 . —J!L==+a J (a? -xy (fVd 1 -x* t ' ^ x V2ax -"?" a« 01 /V ~ 2 )^ 2 6 , *" A 1 LAV 33 '/K3^- log(, 2 + 2, + 3) + a 33. jT^=ll& - 2 log (2s + 3) + 0. n t r dx , . /Sic — 2\ , ^ 35. /"— £L=. $ log (2x + f/f+4?) + a 36./., **,-„■ log(>-24-*V-4z+13)4-C. GENERAL DIFFERENTIATION. 81 __ P dx . _ x 2x - 3, 38. / sin -1 — -=- + (7. ^ Vl + 3x- x* Vl3 39. / -i — -— . -^ tan" 1 =r -f C. «/ x 2 — 6x -f 11 f ^ 4/ a 40. / (z 2 -2z+2)(£-l)d£. -f. /oXCtX a , — — — . v V125. 2tV -4 2 /^dx 43. / 3 sin 2 x cos xdx. sin 3 x -f- C. 44. / 36(a — & cos 2 <&)* sin jc cos 2^. (a — b cos 2 a;)*-f (7. 45. /- 4— dx. log (tanse + aO+C. «/ tan a; -f a & v 46. / (tan x -f- cot xfdx. tan # — cot x -{- 6". 47. Find f| when y 2 - 2xy + c = 0. y^ ^. «# (y x) 48. Given y 2 — 2axy -f a; 2 = c, to find —. - ^. ttiC lx^ (XX j 49. In #^ 2 y + ^ 2 + d x * — 0; change the independent vari- able from x to y. 6f x ^a; 3 dx __ dy 2 ~ dy 3 ~ dy ~ 50. Change the independent variable from x to z in (2a - l)«g + (2* - l)g = By, where to =1 + •. 51. If «/ 2 = sec 2x, prove that y ~ = 3# 5 — ?/. 82 DIFFERENTIAL AND INTEGRAL CALCULUS. 52. Given s = f — hf + 6t; find the velocity (v) and its rate of change (a') when t = 10. v = 206; «' = 50. 53. In t seconds after a body leaves a certain point the rate of change of its velocity is Qt — 12 feet per second; required its velocity and distance from the point of starting. v = 3f - 12*; s = f- 6t\ 54. In the last example how far will the body travel in 10 seconds ? _ ^ + [s f« = 32 + 432 = 464 (ft>)# 4 55. How many times faster is x increasing than log x, when x = ?i? u times. 56. Required the value of x at the point where the slope of the curve y — tan x is 2. 7t T' 57. A man is walking on a straight path at the rate of 5 ft. per second; how fast is he approaching a point 120 ft. from the path in a perpendicular, when he is 50 ft. from the foot of the perpendicular ? l^f ft. per sec. 58. A vertical wheel whose circumference is 20 ft. makes 5 revolutions a second about a fixed axis. How fast is a point in its circumference moving horizontally, when it is 30° from either extremity of the horizontal diameter ? 50 ft. per sec. 59. A buggy wheel whose radius is r rolls along a horizontal path with a velocity v' ; required the velocity (-=7-) of any point (x, y) in its circumference; also the velocity of the point hori- zontally {jij and vertically (-77). The curve described by the point in the circumference of the y wheel is- a cycloid whose equation is x = r vers -1 - — \try — if\ differentiating this and dividing by dt, we have dx _ y dy dt Y2ry - y' 2 dt K) GENERAL DIFFERENTIATION. 83 Again, the abscissa of the center of the wheel is r vers" 1 —\ differentiating this and dividing by dt, and we have ,' = — ^ f (*) V2ry -ifdt Again, since ds 2 = dx* -f- dy 2 , we have ds_ _ J (dxV (dy_ dt~ y \dtl ~r\dt 1 + [± (3) From (1), (2), and (3) we readily obtain 60. In the last example find the values of -^— , —-, and ^- at r dt dt dt the point (1) where y = 0; (2) where y = r; (3) where ?/ = 2r. 61. Water is poured at a uniform rate into a conical glass 3 inches in height, filling the glass in 8 seconds. At what rate is the surface rising (1) at the end of 1 second? (2) At what rate when the surface reaches the brim ? (1) \ in. per sec. (2) -J in. per sec. CHAPTER V. SERIES, DEVELOPMENT OF FUNCTIONS, AND INDETERMINATE FORMS. SERIES. 113. A Series is a succession of terms following one another according to some fixed law. If the sum of the first n terms of an infinite series approaches a definite limit as n increases indefinitely, the series is Conver- gent ; if not, it is Divergent. The sum of a finite series is the sum of all its terms; and the sum of an infinite convergent series is the limit which the sum of the first n terms approaches as n increases. An infinite divergent series has no definite sum. 114. To Develop a function is to find a series, the sum of which shall be equal to the function. Hence the development of a function is either a finite or an infinite convergent series. For example, (x + l) 3 = x* + 3.i' 2 + 3x + 1. This finite series is the development of the function (x + l) s for any value of x. Again, by division we obtain = l-fa; + ^-f^ + ... x n '\ ... (1) 1 -x Now this series is the development of only for values 84 SERIES. 85 of x numerically less than 1, for the omitted remainder is x h 1-x ; hence, denoting the series by s, we have 1 x r ' l-*~* + l-af 1 x n Therefore s can be the value of only when = 0, 1 — x J 1 —x ' and s the development of only for such values of x as will x — X x n cause to approach 0, as n increases indefinitely, and this _L — X can be the case only when x < 1. Thus : (1) For x = 2 we have -1 = 1 + 2 + 4 + 8 + ... 2"- 1 - 2% which would be absurd were the remainder — 2 n omitted. (2) For x = i we have 2-i + I + L , I . J_ + l in which the remainder — decreases as w increases, indefinitely. 115. A series is said to be absolutely convergent when it remains convergent on making the signs of all its terms positive; but only conditionally convergent when it becomes divergent on such a change of signs. The series 1 — i + 3 — i + . . . is an example of a condi- tionally convergent series. 1 16. Prop. TJie infinite series u x + w 2 + . . . u n _ x + u n + . . . will be absolutely convergent if the terms are all finite, and the limit 11 of the ratio — - , as n is indefinitely increased, is 7iumerically less than unity. 86 DIFFERENTIAL AND INTEGRAL CALCULUS. u Since the limit of — - -, as n is indefinitely increased, is less than 1, there must exist some finite integer, c, such that for all values u of n which are greater than c, — — is less than 1. The sum of the terms u x + ? ' 2 ~\~ • • • u c is a definite finite quantity, and it only remains to show that the sum of the remain- ing terms, u c+1 -\- u c+2 +.-...., is also. Let 7 be a number less than 1 but greater than any of the ratios 5**, %&, ^±2, . . . ; then u c u c+l u c+2 U c ^i k the series is divergent, and when x = k the series in some cases is convergent, and in others di- vergent. 119. Cor. III. The series f'{x) = a, + 2a 2 x + 3fy? 2 -f . . . (n — l)a n _ x x n ~ % -f- ?ia n x n - 1 , obtained by differentiating /(») = a -f- «,£ -f ft 2 ^ 2 H~ • • • a n-iZ n ~ 1J r a n % n , is convergent for the same values of x as the last-mentioned series. For in the former k — the limit of '--^. which is the na n same as the limit of — — . a n It is also evident that the limits of convergence of the series obtained by integrating the individual terms of f(x)dx are the same as those of the series f(x) itself. EXAMPLES. Find the values of x which will render the following con- vergent : x 9 x a x"- 1 x n , Here a n _ x = and a n = -. .*. — = = 1 A , n — l n a n n — 1 n— 1 which = 1 when n = co . Hence (117), — 1 < x < 1; that is, x lies between — 1 and -f- 1. „ , , , z 2 , a; 3 , x* X"' 1 x n , Here -^ = - — =—— = n. which = co when n = oo ; hence a n \n — 1 ' -co < a; • • W or f(y + x ) = f( l j)J r Ax^m,x% (2) where A (= m l -=f f (y) ) is a constant with respect to x, and mj? is the acceleration of u (Art. 25), and this is what we wish now to determine. Since m n is of such a character that m^x vanishes with %, we will assume (see footnote, p. 10) m 2 = B + Cx + Dx* + . . . Lx n ~ 3 , .... (3) where B, C, D, . . . L are independent of x, and x has such a value as to render the series (if infinite) convergent. Substitut- ing in (2), we have f(y + x)=f(y) + Az + Bx*+Cx> + ...Lx n -\ . (4) DEVELOPMENT OF FUNCTIONS. 89 Differentiating (-1) successively with respect to x, we have f'{y + x) = A + 2Bx-\-3Cx' + ...(n-l)Lx n - 2 ; (5) f(y + x) =2B + 6Cx + ...(?i-l){n-2)Lx n - 3 ; (6) f'iy + X ) =QC + ...(n- \){n - 2){n - 3)Zz n " 4 ; (7) • • J f n - 1 (y + x) = \n-lL (8) These equations, (5), (6), (7), etc., are true for any value of x which renders equation (4) convergent (Art. 119) ; therefore they are true when x = 0, which gives f'(y) = A, .-. A=f'(y); f"(y) =22?, .:B = ^f"(y); f'"(y)=eO, ... C=i-/'"G/); f-\y) = \ n-l L, ,-. L = -—j-/-^). Substituting these values for A, B, C, ... L in (4), we have (A) This is the formula required, which was first published in 1715 by Dr. Brook Taylor, from whom it takes its name. The preceding is not a rigorous demonstration of Taylor's formula, inasmuch as the possibility of development in the proposed form is assumed. A rigorous proof, including the form of the remainder, has been inserted in the Appendix, A. , to be used or not, according as the teacher or student may desire. 90 DIFFERENTIAL AND INTEGRAL CALCULUS. 122. Cor. I. To determine for what values of a; the series is convergent. The nt\\ and (n -f l)th terms of the series are evidently /"W^T and /%)£. Therefore (Art. 117) the series is convergent for any value of x numerically less than jpvi + fM, or JfTM), when „ =00; \ n-i \n \ r\y) r f n ~l(y) Hence, if • . , / is not zero when n =oo , the series is con- f{y) f~(y) vergent for all finite values of x; and if n ~-j^ ( ^r — 0, when n = oo , the series is divergent for all values of x except 0. In deducing Taylor's formula we have supposed all the func- tions that occur to be continuous. Hence the formula is inap- plicable, or "fails," if the function, or any of its differential coefficients, be infinite for values of the variable lying between the limits for which the development holds. 123. To develop (y -f x) m . Here f(y + x) = (y + z)~ Make x = 0, f(y) = y m . Differentiate, etc., f'(y) — my m ~ 1 , /"(y) = m(m - \)y m ~\ f"\y) — m(m - l)(m - 2)y m ~\ etc. etc. Substituting these values in (A), we have (y -f x) m = y m -f mxy m ~ l -f v |9 f x*y m ~* , m(m — l)(w? — 2) 3 m . , ,_. + — r^~ L x 3 y m -\ etc., (B) which is the Binomial Theorem. DEVELOPMENT OF FUNCTIONS. 91 Cor. I. Let us determine for what values of x the equation is true,, supposing m negative or fractional. f n ~ l (y) = m(m — 1) . . . (m — n + 2)tj m - n+1 , f\y) = m(m — 1) . . . (m — n + l)y m - n ; ... (Art. 122), n^ {y) - f\y) m -n + ¥ which = — y when n = go . Therefore formula (B) is true when x is numerically less than y. Cor. II. Making y in (B) equal to 1, we have (l+x) m =l+mx+ — ^ V+-* ^ V + etc., (C) in which — 1 < x < + 1- 124. To develop sin {y -\- x). Here f(y -\- x) = sin (y -{- x). Making x = 0, and differentiating, we have f{y) = siny; /'(#)= cos y; f"(y) = - sin y; f"\y) = - cos y ; f iv (y) = sin y ; etc. Substituting these values in (A), we have sin (y + a) = sin y (l - -|- + ~- -— |- + etc.\ L y f (D) / ar . ■ ar z 7 , , \ + cos ^f-|3"+|5"-r+ etc - Cor. I. In (D) by making y = 0, remembering that sin = and cos = 1, we have /y 3 /£ 5 /v T sin x = x _—+-—_ — -j- etc. ...(E) 92 DIFFERENTIAL AND INTEGRAL CALCULUS. Coe. II. Differentiating (E), we have x* x i x* a »s = l__+___ + cta. . . . (F) Cor. III. For the quantities within the parentheses in (D), substituting their values from (E) and (F), we have sin (y -f- x) = sin y cos a; -j- cos y sin #. . . . (G) Coe. IV. Differentiating (G), regarding a; as constant and «/ as variable, we have cos (y -{- x) = cos y cos # — sin y sin z. . . . (H) 125. To develop log (y -\- x). f(y -\- x) = log (y -\-x); making x = 0, and differentiating, we have M = 1^ (y)> f'(y) = J-; /'%) = - £; /"%) = | s ; /'%) = - |r, etc. Substituting in (A), we have log (y + *) = log (y) + ^ - ^5 + Jp - etc., . . (I) which is the logarithmic series. Coe. I. The nth. and (n + l)th terms of (I) are, omitting xi • x *~ l ^ x " i a j. Vim a " _1 w y the signs, -. . - , , and — - ; hence, Art. 117, = — ^-, to (n — l)y l ny n a n n — 1 which = y when n == go . Therefore formula (I) is true when # is numerically less than ?/. Coe. II. In (I), by making y = 1, we have Iog(i + *)=«-£ + y-5 + «to.. ■ • (K) which is true for all values of x numerically less than 1. DEVELOPMENT OF FUNCTIONS. 93 126. Maclaurin's Formula is a formula for developing a function of a single variable, as y = a x , y — log (1 -\- x), y = (« + z) n . It may be derived from (A) by making y = 0, which gives f(x) = f(0) +f(0)x +/"(0) * + /"'(0)4 + --/- 1 (0) 1 ^ I + ..., (L) in which /(0),/'(0),/'"(0), etc., represent the values which /(a:) and its successive derivatives assume when x = 0. Coe. I: Formula (L) is true for all values of x numerically less than ^ , when w = x . 127. To develop a x . Here /(z) = a x , .-. /(0) = a = 1; /'(&)= a* log a, .\/'(0) =loga; f"(x)=a*log'a, .-. f'(0) =log 2 a; /'"(*) = «* log 3 a, .-./'"(()) = log 3 a; etc. etc. Substituting these values in (L), we have a* = 1 + log ax + log 2 a— + log 3 a-^r + log 4 a-^-, . (M) which is called the Exponential Series. Cor. I. This series is convergent for all finite values of x, since /" _1 (0) -^/ n (0) is obviously finite and different from zero for all values of n. Coe. II. Making a = e, remembering that log e = 1. we have ^l + s + ^ + i^ + JjL + etc.. . . (N) 94 DIFFERENTIAL AND INTEGRAL CALCULUS. Cob. III. Making x = 1, we have Hence e = 2.718281 +. 128. Find the development of tan -1 x. In the applications of Maclaurin's formula the labor of find- ing the successive derivatives can often be lessened by taking the development of the first derivative, as follows: /0) = tan" 1 ^ /./(0) = tan- 1 = 0: f ' {x) = Ih^ 2 = * ~ ^ + x * " x * + etc " ■'" /r(0) = * f'(x) = - 2x + 4x* - 6x b + etc., .\ /"(0) = 0: f"{x) = -2 + 3. 4. x 2 - 5.6z 4 + etc., -•./'"(O) = -2; /%) = 2 . 3 . 4:X - 4 . 5 . 6x 3 + etc., .*. f y (0) = °l f(x) = |4 - 3 . 4 . 5 . 6x 2 + etc., .'. / v (0) = |4 ; f*(x) = - 2 . 3 . 4 . 5 . 6£ + etc., ,\ /^(O) = 0: /^(a.) = _ |6_ + etc., .-. fi*+77^ + etC -J DEVELOPMENT OF FUNCTIONS. 97 Taking six terms of this series, we have log 2 = .693147+ . . . Patting z = 2 in (4), we have log (3) = log 2 + 2(1 + 3JL + _L + _L + etc.) = 1.098612 + log (4) = 2 log 2 = 1.386294 +. Putting z = 4, we have log (5)=log4 + ^ + ^ + ^ + ^- 7 + etc.) = 1.609437 +. log 10 = log 2 + log 5 = 2.302585 -f . In this way we can compute the natural logarithms of all numbers. It is not necessary to use the formula in finding the logarithms of composite members, for they can be found by simply adding the logarithms of their factors. Thus log 15 = log 3+ log 5. 131. To compute common logarithms. The modulus of the common system is m = log 10 e (Art. 79). Hence 10™ = e, .'. log (10 m ) = log e, or m log 10 = 1. Let log 10 v = n and log v = n'; then 10 n = v and e n ' = v; 10 n = e n/ , log (10 n ) = log e n ', or n log 10 = n\ .-. n = (n') .434294 +. Hence, to find the common logarithm of any number we multiply the natural logarithm of that number by the modulus of the common svstem. 98 DIFFERENTIAL AND INTEGRAL CALCULUS. INDETERMINATE FORMS. 132. In algebra — is called a symbol of indeter ruination, since any number whatever may assume this form. Thus, wXO = 0: divide both sides by and we have n — -. There are many fractions which assume the form of — in consequence of one and the same supposition, which makes both numerator and denominator = 0. Snch fractions are called Vanishing Fractions, and their values, which appear under the form of — , can generally be determined by the calculus. /£ 3 d" o Thus the fraction -= becomes — when x = a. x — a This form arises from the existence of a factor (x — a) com- mon to both numerator and denominator, which factor becomes under the particular supposition. Dividing both terms by this factor, we have x s — a 9 x* + ax 4- a* , . , 3a . , which = — when x = a. sf-a* x + a ' 2 133. To evaluate a fraction that takes the form of - Let u and v be functions of x such that when x = a, u — 0, and v = 0. Let u and v be estimated from the point where their values are 0, that is, from where x = a; then when a: (= a) is increased by h we shall have, identically, u _ Au v ~ Av INDETERMINATE FORMS. 99 As h approaches 0, or x approaches a, the limit of — is du equal to -=— (Art. 27) ; therefore u "1 _ ^w which is read, when x = a, — is equal to — . Applying this to the preceding example, we have x 3 - a s ~\ _ d(x* - ft 3 ) ~l _3aH _3« z* - a 2 ] a ~ d(x* - d 2 )_\ a ~2zJ a ~ 2' An easy deduction of the rule for reckoning forms like — is obtained by the use of Taylor's formula, as follows : Let (a 4- h)~~\ _ limit [- 0(g) + 0»ft + |0'»A 2 + . .."] _ "(a) = 0, and /"(a) — 0, we have ; and so on. 0"'(a) _ 0(g)" Hence, Rule. Substitute for the numerator and denom- inator, respectively, their first derivatives, or their second derivatives, and so on, till a fraction is obtained whose terms do 100 DIFFERENTIAL AND INTEGRAL CALCULUS. not both become ivhen x = a ; the values thus found will be the true value of the vanishing fraction. EXAMPLES. Compute the following: x* — 16 1 8 - 20j, x' + x - 20j; 9 5x* — 8x + 3~l 2 a- 7z 2 -9z + 2J 1 5 z 3 - 3z + 2 ~| 3 a- x 3 -x* -x + IX 2 ■]• _Jo _, 1 — COS X , 5. : . 0. sin # a - V~aF^-~x~*^ « Jo 9. 10. sm x 12. -u- sing-g cos af] 2a fi ^ — sin af| « 3 ' in' ] - 1 log sin x ~| 1^ (^-2^.* 8 1 6 sin 3x ~1 3 a; — | sin 2£_J * ~ 2 e x — e~ 2 sec 2 x — 2 tan aTl 1 1 + cos 4z X 2 4 a — sm a? J (e*~l) 3 J/ 6" INDETERMINATE FORMS. 101 There are other indeterminate forms besides -, such as — , oo oo x 0, co -co, 0°, go °, 1°°, which will be considered in succes- sion. 134. To evaluate a fraction that takes the form of — . 00 Let u and v be functions of x such that u =co and v =oo when x = a; then for the same value of x, - = and - = 0. Hence 1 u v , — = — = -, when x = a. V 1 ? u Art. 133, - = — — - = -r=-, when a; = «. . . . (1) v /1\ v^w' v 7 Dividing (1) by — , we obtain udv . w~| _ £?w"| Now (2) is derived from (1) by dividing by — ; hence, if — is finite, (2) is true for all finite values of — ; and if - = or oo , it v v may be shown that (2) is true in these cases also. it Suppose ——0 when x = a, and k a finite quantity, u . 7 u + lev 7 then — \- k = = k. v v To this last fraction (2) evidently applies, hence u + kv du + kdv u , , du , , — = ~ , or - + k = -=- - + k\ v dv v dv ,, , . u du , that is, — = -^— , when x = a. v dv 102 DIFFERENTIAL AND INTEGRAL CALCULUS, 11 v If — — co , then — = 0, and we have the preceding case. Therefore the form — is to be evaluated in the same way as the form -. EXAMPLES. Find the values of the following; i. * n 2. d(x) ~ d(\ogx) ax* -}- lr ex* -f- d_j 3. ^=S\ . X J 5. 6. i=n =*i tan x log x~ 1 1- log tan 2x "1 log tan x J*" log cot a; cosec x tan^-(z + l)j 7TX tan — 9. log cos (inx) log (1 - x) 1 a c ' CO. 0. 0. 0. 1. INDETERMINATE FORMS. 103 135. To evaluate a function that takes the form of X °o or oo — oo . Functions of this kind can be transformed so as to assume the form of - or — , and then be evaluated by the previous methods. EXAMPLES. Find the following: 5 1. sec 3x cos bx] . — -. This takes the form of oo X 0; but sec 3rs cos bx cos bx cos 3x' which takes the form of . 2. sec x — tan x] r . 0. This takes the form of co — oo ; but sec x — tan sma; 1 — sin x . = , which takes the form oi -. cos x cos x 1 log )gx a — lj, ? X s . 4. cosec e x — e x 6. (1 — tan x) sec 2x]„ . 1. cos # „ x — a\ 7TX 7. — tan — - a 2a 8. x sin a; _ a- aT 9. # m log n x\. (m and ft being -j-.) 0. 2 10. (1 - • x) tan (Itt^)],. 7t 104 DIFFERENTIAL AND INTEGRAL CALCULUS. 136. To evaluate a function that takes the foim of 0°, co °, -I 90 or 1 . Take the logarithm of the given function, which will as- sume the form of X co , and can be evaluated by Art. 135. From this the value of the function can be found. EXAMPLES. Find the following: This takes the form of l 00 . Making y = (l -J- - ) , we have log y = x log (1 +"—')• The value of x log f 1 -j — ) is found by Art. 135 to be a. Hence, when x = oo , log y = a; .'. y = e a , 2. Vx]^, or x*J m . 1. 1 1 This takes the form of oo °; the log of x x is -logx, the value i is x s 5 ! = i. of which, when x = go , is 0; hence x a 3. (sin z) tan % . 1. 2 4. (cot x) sin *] . 1. 5- if + 1)3.. e. 1 6.(tanf)-?]. 7. (cot a) 1 " 3 ^],. e 1 e ' i 8. ftog (e + *)]„. e J 9. tV + z] . x 10. ^cos 2.r] . ?' INDETERMINATE FORMS. 105 11. fS*] • i. X —I oo 12. (cos^a^X e~ inm \ 137. In implicit functions, as f{x, y) — 0. the derivative -— can be evaluated by the previous methods when it assumes an indeterminate form for particular values of x and y. EXAMPLES. 1. Find the slope of x A — a*xy -\- b*y* at the point (0, 0). TT dy Ix' — cfy , . Here A = te=2 ¥y = WheD X = y = °- a?/ a# ax ^ -^- — ^— = Ti when a; = y = 0: that is, | = * or W®' -<(£)=*£ a 3 — 2^V dx dy . a 2 ,\ / = or ^. £?z £ 2 2. Find the slopes of a 3 — 3azy + # 9 = at (0, 0). and co . 3. Find the slopes of x* -f- ax*y — ay z = Q at (0, 0). and ± 1. 4. Find the slopes of y 2 = x(x + a) 2 at (— a, 0). ± V — a. 5. Find the slopes of x* -f 2«a; 2 ?/ — aif = at (0, 0). and ± 1/2. CHAPTER VI. MAXIMA AND MINIMA. DEFINITIONS AND PRINCIPLES. 138. A Maximum Value of a function is a value that is greater than its immediately preceding and succeeding values, and a Minimum Value is one that is less than its immediately preceding and succeeding values. Thus, while x increases continuously, if f(x) increases up to a certain value, say/(«), and then decreases, f(a) is a maximum value of f(x) ; and if, while x increases, f(x) decreases to a certain value, say/(£), and then increases, f(b) is a minimum value off(x). For example, sin x increases as x increases till the latter reaches 90°, after which sin x decreases as x goes on increasing; that is, sin 90° is a maximum value of sin x. Again, if x increases continuously from to 6,f{x) = x* — 6x -f- 10 will decrease until x becomes 3 and then it will increase; hence /'(3) = I is a minimum value oi f(x), or x* — Qx + 10. Let the student substitute 1, 2, 3, . . . 10, successively, for x in /(a) = x 3 — \%x l + 96z - 20, and thus show that /(4) is a maximum and/(8) is a minimum. 139. Any value of x that renders f(x) a maximum or a minimum is a root of the equation f {x) = or go , if f(x) and f'(x) vary continuously with x. For, if we conceive x as always increasing, f(x) changes from an increasing to a decreasing function as it passes through a 106 MAXIMA AND MINIMA. 107 maximum value, say/(«), and from a decreasing to an increas- ing function as it passes through a minimum value, say f(b). Consequently /'(a) must change sign as x passes through a or b, Art. 25. But f'(x) can change sign only by passing through or oo . Therefore f'(a) or f'(b) = or go ; that is, a and b are roots* olf'(x) = or oo . To illustrate the preceding principles and definitions graphically, let y = f(x) be the equation of the curve Am ; then f'(x) = the slope of the curve at the point P or (x, y), Art. 48. As x (= OB) increases, the point P will move from A along the curve to the right, and y or f(x) will in- crease till it becomes aa', and then °' a B b c de h m ' decrease till it becomes bb', and then increase, etc. Therefore aa\, cc/, ee' are maximum, and bb', dd' are minimum, values oif(x). The slope of the curve, f'(x), is evidently positive before, and negative after, each maximum value of f(x); and negative before, and positive after, each minimum value of f(x). Moreover, at the points where f(x) is a maximum or a minimum, the curve is either parallel or per- pendicular to the axis of x, and therefore/' (x) = or qo , The converse of this theorem is not always true; that is, any root of f'(x) = or oo does not necessarily render f(x) a max- imum or a minimum. It is our purpose now to determine which of the roots will render f(x) a maximum and which a minimum. 140. If the sign of f'{x) undergoes no change as f'(x) passes through or go , the corresponding value of f(x) will be neither a maximum nor a minimum. For, so long as the sign of f'(x) undergoes no change, f(x) * Here, and in what follows, the word root includes the real values of x which satisfy the equations f'{x) = or oo , whether/'^) be an algebraic or a transcendental function. 108 DIFFERENTIAL AND INTEGRAL CALCULUS. does not change from an increasing to a decreasing function, nor vice versa. Cor. I. If an even number of the roots of f'{x) — or oo are equal to a, then x — a will not render f(x) a maximum or a minimum. 141. If the sign off'{%) undergoes a change as f'(x) passes through or oo, the corresponding value of f(x) is a maximum or a minimum. For, if fix) undergoes a change of sign, f{x) necessarily passes from an increasing to a decreasing function, or vice versa. Cor. I. If an odd number of the roots of f'(x) = or oo are equal to a, then x = a will render f(x) a maximum or a minimum. Cor. II. Therefore, omitting the equal roots of which the number is even, every real root of f'{x) = or go will render f(x) either a maximum or a minimum. Now, of these let us find which will render f(x) a maximum and which a minimum. 142. Maxima and minima of a function occur alternately. For, suppose that f(a) and f(b) are maxima of f(x), where a < b. Just after passing through f(a),f(x) is decreasing, and increasing just before it reaches f(o); but in passing from a decreasing to an increasing state it must pass through a mini- mum; hence there is one minimum between every two consecu- tive maxima. Cor. I. Denote the roots oif'(x) — and /'(#) = co which will render f(x) a maximum or a minimum by a lS a 2 , a 3 , a A , etc., in ascending order or algebraic magnitude. Then, if /(#) is an increasing function for all values of x less than a lf that is, if f'(a 1 — h) is positive, h being ever so small, /(a x ),/(rt 3 ), f(a b ), etc., are maxima, and f(a 2 ),f(a A ), etc., are minima; and if f(x) is a decreasing function for the same values of x, that is, if f'(a x — h) is negative, the maxima and minima will be inter- changed. MAXIMA AND MINIMA. 109 RULES FOR FINDING MAXIMUM AND MINIMUM VALUES OF FUNCTIONS. 143. The preceding principles indicate the following rule for finding the values of x which will render any function as f(x) a maximum or minimum: Differentiate the function f(x); make /'(%) = and f'(x) = qo ; find the real roots of both equations, and arrange all of them in order of algebraic magnitude, as a 1 , a. 2 , a 3 , etc., omitting the equal roots when there are an even number of them; substitute — oo or a l —h, li being very small, for x in f'(z), and (1) if the result is -J-, a i9 a 3 , etc., will each render f(x) a maximum, and a 2 , a A , etc., will each render f(x) a min- imum; (2) if the result is —,/(rt 1 ),/(rt 3 ), etc., will be minima, and /(ff a ), /(aj, etc., will be maxima. 144. The preceding rule requires that all the real roots shall be found; it is sometimes desirable to know independently whether any particular root as a' will render f(x) a maximum or minimum. This may be done thus: I. Substitute a' — h and a' -\- h for x in f'{x), h being a small quantity, and (1) if f'(a' — h) is -f, and /'(a' -f- h) is — , f(a') will be a maximum; (2) if f f (a 9 — li) is — and f'{a! + h) is -\-,f(a r ) will be a minimum, and if f'(a' — h) and/'(&' -\-h) have the same sign, /(a') will be neither a maximum nor a min- imum. 145. II. Developing f(x — h) and f(x + h) by Taylor's formula, substituting a' for x, transposing /(a'), and remember- ing that f{a') = 0, we have /i«' - *) -/(«') =/'>')! -rwj +/ Lv K)|r - (i) t V- li and 110 DIFFERENTIAL AND INTEGRAL CALCULUS. If li be taken very small, the sign of the second member of either (1) or (2) will be the same as the sign of its first term. Hence, if /"(«') is negative, f(a') is greater than both f(a' — 7i) and f(a r -(- li), and therefore a maximum; while if /"(«') is positive, f(a') is less than both f{a' — li) and f{a' -f h), and therefore a minimum. If /"(«') = 0, and /'"(«') is not 0,f(a') is neither greater than both f(a f — h) and f(a' -\- h) nor less than both, and is therefore neither a maximum nor a minimum. If /'"(«') as well as f"(a') is 0, then, as before, f{a') will be a maximum or a minimum according as f"(a') is negative or po- sitive; and so on. Hence if a' is a root off'(x) = or ho, substitute it for x in the successive derivatives of f(x). If the first derivative that does not reduce to is of an odd order, f(a') is neither a maxi- mum nor a minimum; but if the first derivative that does not reduce to is of an even order, f{a') is a maximum or a mini- mum, according as this derivative is negative or positive. Note. — In many instances this rule is impracticable on ac- count of the great labor involved in finding the successive de- rivatives. 146. The following principles are self-evident, and often serve to facilitate the solution of problems in maxima and minima: (1) If c is positive, f(x) and c X f(x) are maxima or minima for the same value of x; hence a constant positive factor or divisor may be rejected in finding this value of x. (2) logf(x) and/(z) are maxima or minima for the same value of x; hence log may be rejected. (3) c -\-f(x) and f{x) are maxima or minima for the same value of x; hence the constant c may be rejected. (4) If n is a positive whole number, [/(#)]" and f(x) are p maxima or minima for the same value of x; hence in [f(x)] q the denominator q may be rejected, or in Vf(x) the radical may be removed. MAXIMA AND MINIMA. Ill EXAMPLES. 1. Find what values of x will render x 3 — 3a; 2 — 24a; -}- 85 a maximum or a minimum. Here f(x) = x 3 - 3x* - 24a; + 85, f'{x) = 3a; 2 - 6a; - 24, f"(x) — 6x — 6. Making /'(a;) = 0, we have 3z 2 — 6x — 24 = 0, the roots of which are x = + 4, x — — 2. Xow to determine whether these values of a; give maxima or minima values of fix), we substitute them for x in /"(a-). Thus: /""(4) = 6x4-6 = + 18, /"(- g) = 6 X - 2 - 6 = — 18. Hence, Art. 145, when x = 4, /(a:) is a minimum, and when x == — 2, /(a;) is a maximum. Let the student construct the locus oiy = x 3 — 3a; 2 — 24a; -f- 85, and thus exhibit these results graphically. 2. Examine f(x) = x 3 — 3a; 2 + 3a; -f- 7 for maxima and minima. Here /'(.r) = 3z 2 - 6a; + 3, f"(x) = 6x - 6,f'"(x) = 6. The roots of 3a; 2 — Qx -\- 3 = are x = 1, a; = 1. Substituting these values of x mf"(x) and /'"(a;), we have/"(l) = 0,/'"(l) = 6. Therefore the function/(a;) has neither a maximum nor minimum value, which we also infer from the fact that the tivo roots of f(x) = are equal, Art, 140, Cor. I. 3. Find the maxima and minima of x b — 5a; 4 -f- 5a; 3 — 1. Here f'{x) = 5a; 4 - 20a; 3 -f 15a; 2 ; .:f'(x) = is bx* - 20a; 3 -f- 15a; 2 = 0, or (a; 2 — 4a; -f- 3)a; 2 = 0; the four roots of which are 0, 0, 1, and 3. Rejecting the two equal roots, Art, 140, we have a x = 1, a 2 = 3, and since /'(— oo ) = 5(— qo ) 2 is -f, the given function is a maximum when x = 1, and a minimum when a; = 3. Therefore /(l) = is a maximum, and /(3) = — 28 is a minimum. 112 DIFFERENTIAL AND INTEGRAL CALCULUS 4. Examine (x — l) 4 (a;-|- 2) 3 for maxima and minima. Differentiating and reducing, we have The Foots of f'\x) --= are those of (x - l) 3 = 0, (x + 2) n = and Ix -j- 5 — 0; hence there are three roots each equal to 1, two each equal to — 2, and one equal to — f. Rejecting the two equal roots, we have a i = — f and a 2 — 1, and since /'( — co) = (— oo) 3 (— co) 2 ( — 7oo) is -f, f(x) is a max imum when x = — f , and a minimum when a = 1. 5. Determine when J + c (x — a)* is a maximum or mini- mum. By (3) of Art. 146 we may remove b, by (1) c, and by (4) 3; hence we have /(a;) == (x — «) 2 ; ■'•/'(«) = 2 (a;— a), and a;— a = 0, or rtj = ^; and since /'( — qo ) is — J the given function is a min- imum when x = a. 6. Find the maximum and minimum values of f{x) = y- , ( 8 « Here H*) = f±f • I. /'(a?) = gives x(x -\- 3) 2 = 0, of which one root is and the other two are — 3 and — 3. II. f'(x) = co gives (x + 2) 3 = 0, the three roots of which is each — 2. Rejecting the tivo equal roots, we have «, = — 2, a„ = 0; and since /'(— oo) is +, /'(— 2) = co is a maximum value of f(x), and f(0) = $£- is a minimum. 7. If the derivative of f(x) is f(x) = x* - 10a; + 21, what values of x will render f(x) a maximum or minimum. The roots of x* — 10a; + 21 = are 3 and 7; substituting these for x in f"(x) = 2a; - 10, we have/"(3) = - 10 = - 4, and/"(7) = 14 — 10 = -J- 4; therefore /(3) is a maximum and /(7) is a minimum. Find the values of x which will give maximum and mini- mum values of the following functions: MAXIMA AND MINIMA. 113 8. u — x 2 — Sx + 12. x = 4 3 min. 9. u = x 3 — 3x 2 — 24x + 85. x = — 2, max. a; = 4, min. 10. u = x 3 — 3a; 2 + 6.^ + 7 Neither a max nor min. 11. u = 2x 3 — 21a; 2 4- 36x — 20. x = 1, max.. a: = 6, min. 12. w = (a; — 9) 3 (.t — 8)\ a; = 8, max., a; = 8$, min. 2^ — 7# 4- 6 13. « = * — — x = 4. max.* a; = 16, min. a; — 10 . (x 4- 2) 3 14. u — J ~ 2 . x = 3, max , x = 13, min. [x — 6) x 2 4- 3 15. m = --. a; = — 1, max. ' x = 3. min. a; — 1 ' 1 — x -j- x' . 16. w = - — : ^ # = +, mm. 1 -\- x — x ir . (a — x) 3 a iv 2* = -^-. 2;=-, mm. a — 2a; 4 9 4 18. u.= -~\- . a; = 9, max.; a; = 1|, min. a* o — a; 19 w = sin a; { cos x. x = —, max. 4 7t 20. w = sin a; (1 + cos a;), x = - , max. sin a; 7T 21. u = - — —r -. a; = — , max. 1 4- tan x 4 x 22. w = - — : 7 . x = cos a;, max. 1 4- x tan x 23. u = (l + a; 1 ) (7 — x) 2 . x—\, max.; x=0 and x=l, min. 24. If x + y — 7i, what is the greatest possible value of xy ? in'. Make u = xy = x(n — x) = nx — x 2 . 25. If y = mx -f- c, find the least possible value of Vx 2 -\- y 2 c ' Vrri 2 + 1 ' Make u = Vx 2 + # a = ^(1 4- w 2 )z 2 4- 2racrc 4- c 2 . 114 DIFFERENTIAL AND INTEGRAL CALCULUS. 26. A merchant bought a bolt of linen, paying as many cents for each yard as there were yards in the bolt, and sold it at 20 cts. per yard; required the greatest possible profit. $1.00. 27. A club of x members has x 3 — 12ar -f 45a; -j- 15 dollars in its treasury; how much is that apiece if the amount is (1) a minimum ? (2) A maximum ? (1) $13.00; (2) $23.00. 28. Find the value of cp when sin f{x, y), for all small values of h and Jc, positive or negative. 148. Conditions for maxima and minima. — In the function u =f(x, y) if we suppose x and y to vary simultaneously, it is obvious from Art. 139, that the maximum or minimum values of u will occur at the points where the total differential of u, [du], is equal to zero. That is, when [*] = S*+^ = d> As dx (= h) and dy (= h) are independent of each other, each term of (1) must be equal to zero. Hence j- = 0, and ¥ =0 (2) These equations express the first conditions essential to the existence of either a maximum or a minimum. Again, as u passes through a maximum or minimum value, [du] changes from + to — , or — to +, respectively ; therefore, in the former case [du] is decreasing, hence [d*u] is — , and in the latter [du] is increasing, hence [dht] is -f-. But the signs of ^^ =J S dx ' + Sw dxd y+w df - - ■ (3) must evidently be independent of the signs of dx and dy, how- ever large or small these differentials may be supposed to be. This can be the case only when f^u\(£u\ ( d*u y \dx" l\dy* I > \dxdy) w MAXIMA AND MINIMA. 119 For, making ^ = ^-, £ = ^. C = ^ , we have ^- + 8£Ai + OT = ^ + ^ + (^-^ , (5) In order that (5) may preserve the same sign for all values of h and k, it is necessary that AC — B 2 should be positive ; for if negative, the numerator of (5) will be positive when k = 0, and negative when Ah -j- Bk — 0. Hence we have as an additional condition for a maximum or a minimum, AC > B 2 , or (4). With this condition, the sign of (5) depends on that of the denominator A. Hence for a maximum we must have d 2 u A or and for a minimum Ao *^<°> • ( 6 > Aor d^> < 7 ) It should be noticed that AC > B 2 requires that A and C should have the same sign. The exceptional cases, where B 2 = AC, Or where ^4 = 0, B = 0, (7=0, require further investigation, but we shall not consider them in this book. The conditions for a maximum or minimum value of u = f(x, y) are then, viz. : For either a maximum or a minimum, clu _ 3 du /rtX ^ = 0, and ¥=0; . .... (8) , d 2 u d 2 u I d 2 u \ also d*df > \d^) (9) For a maximum, — < 0, j-~ < 0. ..... (10) For a minimum, ~^~ > 0, — > (11) 120 DIFFERENTIAL AND INTEGRAL CALCULUS. EXAMPLES. 1. Find the minimum value of u — x* -j- y 3 — 3axy. Here ^ = 3a; 2 - Zay; p- = 3y 2 - Sax; d\i d" 2 u d 2 u _ also ^=e.r, w = *y, a^/ = - 3a - Applying (8), we have x 2 — ay = 0, and y* — ax = ; whence x = 0, y = 0; and x = a, y = a. The values x = 0, y = 0, give cPu _ i% _ d 2 w dx* dy 2 dxdy which do not satisfy (9). Hence they do not give a maximum or a minimum. The values x = a, y = a, give d*u „ *? 8 w . d'u which satisfy both (9) and (11). Hence they give a minimum value of u, which is — a 3 , 2. Find the minimum value of x* + xy -f # 3 — ax — by. $(ab — a 7 — b*). 3. Find the maximum value of (a - x){a - y)(a + y - a). -^ 4. Required the triangle of maximum area that can be inscribed in a given circle. The triangle is equilateral. 5. Divide a into three parts, x, y, a — x — y, such that their continued product, xy(a — x — y), may be the greatest possible. x = y = a-x — y = -. MAXIMA AND MINIMA 121 6. Divide 45 into three parts, x, y, 45 — x — y, such that #y (45 — x — yY may be a maximum. x = 10, ?/ = 15, 45 — x — ^ ■ = 20. 7. Find the least possible surface of a rectangular parallelo- piped whose volume is a 3 . Qa'\ 8. Find the dimensions of the greatest rectangular parallelo- x* if z 2 piped that can be inscribed in the ellipsoid t + ?r + ^ = 1. CHAPTER VII. APPLICATIONS OF THE DIFFERENTIAL CALCULUS TO PLANE CURVES. TANGENTS, NORMALS, AND ASYMPTOTES. 149. Equations of the Tangent and Normal. In Fig. 23 let P(x lf y t ) be the point of tangency of the tangent TP; then the equation of TP is y — y \ = m(x — o^), where m is the tangent of the angle BTP. But, Art. 26, tan BTP = ^; therefore the equation of the tangent PT is y-^ = ^(*--*i)> ...... (A) cly cly where -^ is the value of ~~ with respect to the curve AP at ax 1 ax L the point (x x , y t ). Since the normal PN is perpendicular to the tangent or curve at P, its equation may be obtained from (A) by substitut- dx x , dy } , . , . mg — -=— s tor -p 1 , which gives y \ i y-y>=- j^( x - *J ( B ) EXAMPLES. 1. Find the equations of the tangent and normal to the pa- rabola ?/ 2 == ±ax. 122 DIFFERENTIAL CALCULUS AND PLANE CURVES. 123 dy _ 2a m dy 1 _ 2a xxere ~^ — — » • • ~^ — • ax y dx x y l dy Substituting this value of -— in (A) and (B), we have V- V, = —(*—»,)* tangent; . . . . (1) y i y - y 1 = - |^(« - *,), normal. ... (2) 2. Find the equations of the tangent and normal to the pa- rabola ?/ 2 = 18ic at the point x x = 2. Here 4# = 18 and y* = 18^; .*. 2a = 9 and #, = 6. Substituting in (1) and (2), and reducing, we have 2y = 3x -f- 6, tangent, and 3y = — 2x -f- 22, normal. Find the equations of the tangent and normal to the follow- ing curves : 3. The circle, %f + x* = R\ (1) yy x + xx, = B>; (2) yx t - xy x = 0. 4. The ellipse, a 2 y 2 + &V = a*b\ (1) a*yy x + Vxx x = a'b*; (2) y-y x = ^(x - x x ). 5. The cissoid, y'\2a — x) = x*. (1) tangent, y-y x = ± a ~ %l) ** (x - x t ). (2a - xj* 6. Find the equation of the normal to y 1 — 6x — 5 at y x — 5. y = - ¥ + ¥• 124 DIFFERENTIAL AND INTEGRAL CALCULUS. 7. Find the equation of the tangent at y x = 4 to the cycloid x = 10 vers- 1 ^ - V20y-y\ y = 2x+ 20(1 -vers ^f). 150. Length of Tangent, Normal, Subtangent, and Sub- normal. Let PT represent the tangent at the point P(x x , y x )> PA" the normal; then y l = FB, BT is the subtangent, and BN\s the subnormal. Let = the angle BTP, then T A B Fig. 23. that is, (2) that is, (3) that is, (4) that is, n (i) BT= BP cot = y 1 dx l subtangent = y t dx x dy' x (0) BN = BP tan BPN = BP tan 0; subnormal = y J dx. TP - {BP + sin 0) = BP ds x ¥7 ; tangent = ?/, ^ fl II A-X 1 CJC Since -^ = — ^ — , -^ = — ] a ' , which substituted in dx y 2 dx 1 y; (K) and (L) give (1) x = x 1 - x x " r which = — 2 when x x — oo . Ax* + a?. (2) y, - y. fi . - t-jj + Ai» which = 2 when x x — oo . DIFFERENTIAL CALCULUS AND PLANE CURVES. 129 Therefore the straight line whose x and y intercepts are — 2 and -1- 2, respectively, is an asymptote to the curve. Since the asymptote passes through the points (— 2, 0) and (0, 2), its equation is y = x -\- 2. 153. General Equation of the Asymptote. Since the asymptote passes through the points (x , 0) and (0, y ), its equa- tion is y = dt {x ~ Xo) ' ' • " (M) ' or y = at x + *•■••■ W This equation enables us to determine whether or not any given curve has an asymptote, and, if it has, to find its equation. Let us denote the values which -^-- and y„ assume when x t = x by m 1 and &, , respectively; then we have y = m x x -f- ~b l (P) 154. When the terms of the equation f(x, y) = are of different degrees, to find the relation of y to x when they are infinite, we may omit all the terms except the group which are of the highest degree with respect to x and y. Thus, when x is infinite, the equation ay* — ~bx 2 -\-cy-\-dx = e gives aif — Ix 1 = 0, or y = ± y —x. ct A curve like a 2 y 2 -j- b 2 x* = a 2 b 2 , or y* = x*(a' — x 2 ), etc., which has no infinite branch or branches, has no real asymptote; this is indicated by the fact that when x is infinite, y, as deter- mined above, will be imaginary. 2, Find the asymptote of the hyperbola a'y 2 — Vx* = — a~b 2 . When x = x , y = ± -x, or y, = ± -x.. 3 a Jl a l 130 DIFFERENTIAL AND INTEGRAL CALCULUS. dx^a'yj y °-V> a* yi - y,' m 1 = ± - and ^ = 0, which substituted in (P) gives y — ± -x, Ans. 3. Find the asymptote of the parabola y* = iax. When x = co , y = ± 2 Va#, or y l = ± 2 Vax y . dy 1 2a 2ax. . — _-.■ , _ jz- = ~ > Vo=y 1 - —r 1 = VfMx , which = co when x,= oo . Therefore the parabola has no asymptote. 4. Find the asymptote of y 3 = ax 3 -\- x 3 . When x = go , we have y 3 = x 3 ; .'. y = x or y x = x t . dy y 2ax, 4- 3a:, 2 .. ax* ~d^ = 3y ,' ' and y » = 3y; ; hence m 1 = 1 and 5, = Q , and the asymptote is y — x -}- ^- . 155. Asymptotes Determined by Inspection. When an asymptote is perpendicular to the axis of x or y, it can often be dx determined by inspection. In the first case m l = oo , or -=-' — 0, which, substituted in (M), gives x — x 1 = 0, since, in this case, x — x 1 ; that is, if y , is infinite when x x is finite, x — x l = is the equation of the asymptote. x % Thus, in the cissoid, y 2 = , ' J 2a — x y = oo when # — 2a; hence the line a; — 2a — 0, which is paral- DIFFERENTIAL CALCULUS AND PLANE CURVES. 131 lei to the axis of y and at a distance 2a from it, is an asymptote to the curve. Again, in xy = a or y = — , when x = 0, y = co ; therefore x x — 0, or the axis of ?/, is an asymptote to the curve. Similarly, in y — a x , when x — '— oo , y = 0; hence # = 0, or the axis of x, is an asymptote to the logarithmic curve. 5. Find the asymptotes of xy — ay — bx — 0. (1) x - a = 0; (2) y - b = 0. 6. Find the asymptote of y 3 = ax 2 — x\ i a a 7. Find the asymptotes of y — c -f- y — c and x = b. 8. Find the asymptotes of y 2 (x* + I) = x 2 (x 2 — 1). y=±x. 9. Find the asymptotes of y*(x — a) = x 3 -\- ax 1 . x — a and y = ± (x -\- a). CURVATURE. 156. A point moving along an arc of a curve changes its direction continuously, and the total change of direction is called the Total Curvature of the arc. The angle TtP', Fig. 25, through which the tangent PT rotates as the point of tangency P moves from P to P', being 132 DIFFERENTIAL AND INTEGRAL CALCULUS. the total change of direction of the point P, is the total curva- ture of the arc PP'. 157. Uniform Curvature. The curvature is uniform when, as the point of tangency moves over equal arcs, the tangent turns through equal angles; that is, when the distance described by the point varies as its direction. Let APm be the curve, AP = s, PP' = As, XEP = 0, Art. 49; then TtP' = Acp. Let PC and P'C be normals meet- ing at C. Supposing As oc Acp, we have (Art. 12) A \ Acp As — 7)i Ad), or — = —j- m 7)1 As (1) Let us consider the meaning of —r- . If the distance As gives a total curvature of Acp, since As oc A •■•t > = t ™- 1 %' hence we have d(p = <&+% ' also ' ds = {dx ' + df)i - . p *_ ds _{da? + dtff ( 1+ Ui) m • u - d dx ~ a?y y dx x ~ tfy* ' b* [ aY J ~ a A V At the vertex x = a, y — 0, R = — , and at the vertex x = 0, a 2 y = b, R = j-> which are respectively the minimum and maxi- mum radii of curvature. 3. Find the radius of curvature of the cycloid ■xV x = r vers -1 — — v%ry — y' Here - = —=M= ■ ■ 1 + ^! = ^. dy V2ry - y> ' dx' y » DIFFERENTIAL CALCULUS AND PLANE CURVES. 135 d*y r dx 2 . \ R = 2 f 2ry, which equals twice the normal. 4. Find the radius of curvature of the logarithmic curve y = *". ^ = (m 2 + y 'f my 5. Find the point on the parabola y* = 8# at which the radius of curvature is 7|§. y = 3, x = 1-J. 6. Find the radius of curvature of y = x* — 4x 3 — l&c' at the origin. „ _ 1 K ~ 36 * 7. Find the curvature of the equilateral hyperbola xy = 12 at the point where x = 3. 1 __ 24 R ~125* 8. Find the radius of curvature of the catenary / x - -\ * = £ 9. Find the radius of curvature of the hypocycloid x l + y 1 = a* A' = 3( W = 0i substituting in (2), we have (r 2 + a 2 ) 1 = a(l + fr) 1 r 2 + 2a 3 2 + 2 * Find the radius of curvature of the following: 2. The logarithmic spiral r = a 9 . R — r Vl -\- (log a)' 9 3. The cardioid r = «(1 — cos 0). P = - \ y 2ar. o a a 4. The lemniscate r 2 = a 2 cos 20. R = —~ , 3r CONTACT OF DIFFERENT ORDERS. 161. Let ?/ =f(x) and ?/ — 0(a) be any two curves referred to the same axes. Let the curves intersect at the point P, whose abscissa is a, then f(a) = 0(a). If f{a) = 0(a), and /'(a) = 0'(a), the curves are tangent at P, and are said to have a contact of the first order. If f(a) = 0(a), /'(a) = 0'(a), and f"(a) = 0"(a), the curves have the same curvature at P, and their contact is of the second order. If, in addition, /'"(a) = 0"'(a), their contact is of the third order; and so on. Thus, contact of the nth. order imposes n -f- 1 conditions. 162. Two curves cross or do not cross at their point of con- tact, according as their order of contact is even or odd. Let x = a be the abscissa of the point of contact of the DIFFERENTIAL CALCULUS AND PLANE CURVES. 137 curves y = f{x) and y = x, we have /'(a) = 3s 1 - 6s, 0'(s) = - 3; .'. /'(I) = 0'(1) = - 3; /"(a?) = te - 6, 0"(s) =0; /. /"(I) - 0"(1) = 0; /'"{*) = 6, 0'"(s) = o; .'./'"(I) > 0"'(1). Hence the contact is of the second order. 2. Find the order of contact of the parabola y* = 4z and the line 3y = x + 9. First order. 3. Find the order of contact of the curves y = 3x — x 1 and xy = 3x — 1. Second order. 4. Find the order of contact of y = log (x — 1) and z 2 — 6# + 2y + 8 = 0, at the point (2, 0). Second order. 5. Find the order of contact of the parabola y 2 = 4x + 4 and the circle # 2 + # 3 = 2a; + 3. Third order. 163. Osculating Curves. The curve of a given species that has the highest order of contact possible with a given curve at any point is called the osculating curve of that species. A curve may be made to fulfil as many independent condi- tions as there are arbitrary constants in its equation, and no more. Therefore, in order that y —f(x) may have contact of DIFFERENTIAL CALCULUS AND PLANE CURVES. 139 the nth. order with a given curve at a given point, the equation must involve n -J- 1 arbitrary constants. Hence, as y = ax -J- 5 has two constants, the osculating straight line has contact of the first order. As (x — a) 7 -f- (y — by = r 2 has three constants, the osculat- ing circle has, in general, contact olthe second order. 164. To find the osculating straight line at any point {x' y y') of a given curve y = f{x). The equation of a line is y = ax + b (1) Since the line and curve pass through {x\ y'), we have y' = aaf + b=f&). ...... (2) Also, % = a =/'(*'), ...... (3) since/'(z') = 0' (ab- solving (2) and (3) for a and b, we have .=!£, and t = f-&*. which, substituted in (1), gives Therefore the osculating straight line is a tangent to the curve, as would be inferred. 165. To find the radius of the osculating circle at any point of a given curve, y =f(x). The general equation of a circle whose radius is r is (x-ay + (y-by = r> (1) 140 DIFFERENTIAL AND INTEGRAL CALCULUS. Differentiating twice successively, we have X - a+{ y-b)<^L=0, ...... (2) i + d £+^-^=° < 3 > From (3), 9 -i=-**+J£ (4 ) *»t». * — J^|P* (5) Substituting (4) and (5) in (1), we have By comparing this result with formula 1, Art. 159, it will he seen that the osculating circle is the same as the circle of cur- vature. INVOLUTES AND EVOLUTES. 166. An Involute may be regarded as a curve traced by a point in a thread as it is unwound from another curve, called the Evolute. Thus, imagine a thread stretched around the curve A 1 P i m l with one end fastened at m l ; if the thread is unwound by carry- ing the point at A above and around to the right, that point of the thread will trace the involute APm of which A l P x m l is the evolute. An evolute may have an unlimited number of involutes, for A may be any point on the curve A x m v DIFFERENTIAL CALCULUS AND PLANE CURVES. 141 111 what follows the chief object is to deduce certain prop- erties of the evolute from its involute, or vice versa, and for uniformity the co-ordinates of P (the involute) will be repre- sented by x, y, and those of P x (the corresponding point of the evolute) by x x , y x ; the arc AP by s; the arc AP X by s r ; and the angles of direction of AP and AP X , at P and P x , by and cp x , respectively. 167. Elementary Principles. I. PP X — the arc AP X = s x . II. PP X is tangent to AP x m x at P x , for it has the same direction as the curve at that point. III. The line PP, is a normal to the curve APm at P. 142 DIFFERENTIAL AND INTEGRAL CALCULUS. For, draw TP tangent to the curve AP at P. . (P^ + tPP^^P) 2 , or ( yi -yy + {x - Xi y = Sl \ (i) (#i - y)(dyi - dy) + (x - £,)(dz - dxj = s.ds,. . (2) Again, P X E = P,P sin EPP t , or y x —y = s ip; . . (3) also, PP = P 2 P cos PPP 1 , or a? — as, = - s^ 1 . . (4) Substituting in (2) from (3) and (4), and reducing, remem- ciii doc bering that dx* -j- dy* = t^ 2 , we have -~ = — — ; that is, tan X = - cot ; .-. X = | + 0, or PFX = f + P^FX; hence P X P is perpendicular to the tangent PT, Cor. I. Since sin f f Vl y if 4a 2 4a a These values of x and y substituted in y 1 — ±ax give which is the equation required; hence the evolute is the semi- cubical parabola. Cor. I. The length of the arc of the evolute s 1 may be found by formula (7), Art. 167. 2. Find the evolute of the ellipse ay + 5V = a 2 b\ {ax^f + {h yi f = {a' - ^) § . 3. Find the co-ordinates of the centre of curvature of the cubical parabola y 3 = tfx. _ a" + 15 y* _ a*y-$ij % Xl ~ 6a*y > Vx ~ ' 2a* ' ' 4. Find the co-ordinates of the centre of curvature of the X _x catenary y = ~(e a + e a ). x ^ =x ~\ *V ~ ^ & = 2 ^ —J 5. Find the co-ordinates of the centre of curvature, and the equation of the evolute, of the hypocycloid x* -f y* = a?. Xl = x + 3 Vxjf, y, = y + dj/xY; fa + V x f + fa - V,f - 2a?. 6. Find the evolute of the equilateral hyperbola xy = m 2 . fa + #i) § ~ fa - ^) § = ( 4 "0 § » Note. — First prove that m fm , xV .. m fm x V 1 ' yi 2\x l 7n/' 1 * l 2 \x ml and thence derive the equation of the evolute. DIFFERENTIAL CALCULUS AND PLANE CURVES. 145 ENVELOPES. 169. Let f(x,y,a)=0 (1) be the equation of a curve, a being some constant quantity. If we assign different values to a, we will obtain a series of distinct curves, but all belonging to the same system or family of curves. One of the curves of this family can be obtained by increasing a by h, thus converting (1) into /I*,y,« + A) = (2) If li be supposed indefinitely small, the curves (1) and (2) are said to be consecutive. The points of intersection of the curves (1) and (2) approach definite limiting positions as h approaches 0, and the locus of these limiting positions, as different values are assigned a, is called the Envelope of the system f(x, y, a) = 0. The quantity a which remains constant for any one curve of the series, but varies as we pass from one curve to another, is called the variable parameter of the series. 170. The envelope of a series of curves is tangent to every curve of the series. Let A, B, C be any three curves of the series, A and B inter- secting at P, and B and C at P' . Fig. 28. As these curves approach coincidence, the limiting positions of P and P' will be two consecutive points of the envelope and of the curve B. Hence the envelope touches B. As an illustration see example 1 under the next article. 146 DIFFERENTIAL AND INTEGRAL CALCULUS. 171. To find the equation of the envelope of a given series of curves. The point of intersection of (1) and (2) will be found by combining the equations. Now, subtracting (1) from (2), we have fix, y, a + h) -f{x y y f a) _ A h ~ U {6} When the curves approach coincidence, h approaches 0, and (3) becomes ^«)=0 W Thus, equations (1) and (4) determine the intersection of any two consecutive curves. Hence, by eliminating a between (1) and (4), we shall obtain the equation of the locus of these intersections, which is the equation of the envelope. EXAMPLES. 1. Find the envelope of y = ax -| , a being the variable parameter. iix y — ax -f- — is the equation of a line, as MN, Fig. 29. tt When a receives an increment It, the line takes a new position, say M'N', which intersects the former line at c. As h approaches 0, c approaches p, a point on the locus (APm) of all similar intersections. Differentiating with respect to a, x and y being constants, we have = x -', whence a = ± \ — . y = ± [ Vmx -\- Vmx], or y* = ±mx. which is the equation of a parabola, DIFFERENTIAL CALCULUS AND PLANE CURVES. 147 Let it be observed that the problem is the same as that of m finding the curve of which y = ax -} is the tangent. 0/ Fig. 29. 2. Find the curve whose tangent is y — mx -|- a Vni 2 -J- 1, m being the variable parameter. x 2 -f- ?/ 2 = a 2 , a circle. 3. If a right triangle varies in such a manner that its area is constantly equal to c, find the envelope of the hypothenuse, or the curve to which the hypothenuse is the tangent. Let OA — a, OB = b; then the equation of AB is b -K--1. (l) But ab = 2c, or b = 2c X - _L W - 1 a " h 2c"■ l, (2) where a is the variable parameter. /2cx Differentiating (2), we get a = y , which, substituted in (2), gives ay = ^-, an equilateral hyperbola. 2 148 DIFFERENTIAL AND INTEGRAL CALCULUS. Solve the preceding problem on the hypothesis that the hy- pothenuse is constantly equal to c. ^ + V* — c*> the hypocycloid. Since a normal to a curve is tangent to the evolute of the curve, the latter is the envelope of the successive normals, or the locus of their intersections. 4. Find the evolute of the parabola y 2 = 4##, taking the equation to the normal in the form y = m{x — 2a) — am 9 , m being the variable parameter. 27ay 2 = 4(# — 2a)\ 5. Find the evolute of the ellipse a 2 y 2 + # 2 ^ 2 = a 2 b% taking the equation of the normal in the form oy = ax tan a' — (a 2 — ¥) sin a', where the variable parameter a' is the eccentric angle. (ax)* + (byf = (a' - b 2 f. See Ex. 2, Art. 168, 6. Find the curve whose tangent is y = mx -f- (a' 2 m 2 -f b 2 )$, m being the variable parameter. a 2 y 2 -f- Vx 2 = a 2 b 2 . 7. Find the envelope of the family of parabolas whose equa- tion is y 2 = a(x — a). y = ± \x. 8. Find the locus of the intersections of x cos a -f- y sin a = p with itself as a increases continuously. x 2 -f- y 2 = p 2 - 9. Find the envelope of all ellipses having a common area (7rc 2 ), the axes being coincident. xy — ± \c 2 . 10. Find the evolute of the curve x* -\- y* = a*, the equation of whose normal is y cos a' — x sin a' = a cos 2a f , where a f is the angle which the normal makes with the axis of x. (x + y f _j_ ( x - y f - 2a\ See Ex. 5, Art. 168. 11. Find the equation of the curve, the equation of its tan- gent being y = 2mx -f- in\ where m is the variable parameter. TRACING CURVES. 172. The Rudimentary Method of tracing a curve is to reduce its equation to the form of y = f(x) ; that is, solve the DIFFERENTIAL CALCULUS AND PLANE CURVES. 149 equation f(x, y) = for y, assign values to x, find the correspond- ing values of y, draw a curve through the points thus determined, and it will be approximately the curve required. This process is laborious, and often impossible on account of our inability to solve f(x, y) = for y. The General Form of a curve is usually all that is desired, and this can generally be found by determining its singular or characteristic points and properties, and these are embraced chiefly in the position of certain" turning-points of the curve, the direction of curvature between these points, and. where and how the branches intersect or meet each other. In addition, we may find, by previous methods, where the curve cuts the axes, whether or not it has infinite branches, asymptotes, etc. 173. Direction of Curvature. — The terms Convex and Concave have their ordinary meaning when applied to the arcs of curves. Thus, as seen from some point below, the arcs AB X and OB are concave, and B x and BE convex. 174. A Point of Inflection is the point at which the curve changes from concave to convex, or from convex to concave ; as the points B J} C, B. Principles. — The slope (-r-) of the curve evidently de- creases as the point P(x, y) moves from A along the curve to the right until P reaches B t , and then increases until P reaches (7, etc. Therefore (1) when the arc is concave, ~— decreases as x increases, hence (Art. 25) its derivative -y~r is — ; (2) when ,, dy . . , d*y . . the arc is convex, -f- increases as x increases, hence -3-5 is 4-. ' dx dx 2 Therefore, I. At any point of the curve y =f(x), the curve is concave or convex according as -^~ is negative or positive. 150 DIFFERENTIAL AND INTEGRAL CALCULUS. d?y dv II. The roots of -~ = or oo which will render —• a maxi- ma; 2 dx mum or minimum are the abscissas of the points of inflections. EXAMPLES. Examine the following curves for concave and convex arcs, and for points of inflection. 1. y = x * - 6x + 7. p? = 2. Since ~^~ is +, the curve is convex at every point. 2. y = re 3 - 6a; 2 + 17a; - 6. |^ = 6(a - 2). The root of x — * 2 = is 2, the point of inflection; the curve is concave when x < 2, convex when x > 2. 3. y = x* - 12a; 3 + 48a; 2 - 50. Points of inflection, a; = 2, a; = 4; curve convex when x < 2 and > 4, and concave when 2 < x < 4. — x ~ 2 ^y 2_ y " a; - 3' dx* " (a; - 3) 3 ' Point of inflection at a; = 3; convex when a; > 3, concave when x < 3. 6.y = log(*-l). _ = _-__ = oo gives (a; — l) a = 0, which has two equal roots; hence, Art. 140, there is no point of inflection; curve concave DIFFERENTIAL CALCULUS AND PLANE CURVES. 151 6. Prove that the curve y = — — — - has points of inflection. ct ~\~ x at (0, 0), (a V3, la VI), {-aVd,-ia 1/3). 7. Prove that the witch of Agnesi, x*y = 4a i (2a — y), has points of inflection at (± \a Vo, |«), and is concave between these points and convex outside of them. 8. Find the points of inflection of y = sin 2x + cos 2x. SINGULAR POINTS. 175. The Singular Points of a curve are the turns and multiple points. A Turn in rectangular co-ordinates is a point at which a curve ceases to go (1) up or down, or (2) to the right or left, Fig. 31. and begins to go in the opposite direction. The former, as B, E, F, G, are called y-turns, and the latter, as 0, D, H, x-turns. The x-turns and y-turns evidently occur at the maximum or minimum values of x and y, respectively. 152 DIFFERENTIAL AND INTEGRAL CALCULUS. A Multiple Point* is one through which two or more branches of a curve pass, or at which they meet. A multiple point is double when there are only two branches; triple when only three, and so on. A Multiple Point of Intersection is a multiple point at which the branches intersect (Fig. 32, a). c e Fig. 32. An Osculating Point is a multiple point through which two branches pass, and at which they are tangent (Fig. 32 ? b, c). A Cusp is a multiple point at which two branches terminate and are tangent (Fig. 32, d, e). A -cusp or osculating point is of the first or second species according as the two branches are on opposite sides (Fig. 32, b, d) or the same side (Fig. 32, c, e) of their common tangent. A Conjugate Point is one that is entirely isolated from the curve, and yet one whose co-ordinates satisfy the equation of the curve. For example, in the equation y = (a + x) Vx, if x is negative y is imaginary, yet the co-ordinates of the point (x = — a, y = 0) satisfy the equation. Hence (— a, 0) is a conjugate point. A conjugate point is, generally, the intersection or point of meeting of two imaginary branches of the curve, and may, in exceptional cases, also lie on a real branch of the curve. There are other singular points, such as Stop Points, at which a single branch of a curve stops suddenly, and Shooting Points, at which two or more branches stop without being tan- gent to each other. But as these rarely occur, they are omitted in this book. * See Taylor's Calculus. co ; . dy 0; . ax DIFFERENTIAL CALCULUS AND PLANE CURVES. 153 176. To determine the positions of the singular points of a curve. Let u =f(x, y) = be the equation of the curve, free from radicals. Then (Art. 109) du dy _ dx dx du dy (a) For the z-turns, -f- v ' dx (b) For the ?/-turns, -p (c) For multiple points, ~i~, by definition, has two or more dy values ; hence, since u contains no radicals, ~ must be of the dx form -. Therefore du , du A t=- = and -r- = 0. /XT/, Hence, to find the a;- turns we have u — and ^— = ; to dy du find the y-turns, we have w = and — = 0; and the values of y and x which satisfy all these equations are the co-ordinates of the multiple points. 177. To determine the character of the multiple points of a curve. From the definitions of the multiple points it follows that: I. At a multiple point of intersection -j- has two or more unequal real values. du II. At an osculating point or a cusp y- has two equal values. III. At a conjugate point at least two of the values of -j- are imaginary. 154 DIFFERENTIAL AND INTEGRAL CALCULUS. EXAMPLES. Find the singular points of the following curves. 1. u = x 2 - xy + if - 3 = 0. . du dy (1) x + 2y = 0; du (2) ^ = 2x-y = 0. . . (3) From (1) and (2) we find (2, 1), (— 2, — 1), the co-ordinates. of the #-turns A and A'. Y Fig. 33. From (1) and (3) we find (1, 2), (— 1, — 2), the co-ordinates of the y-turns B and B'. Since neither pair of these values satisfies (1), (2), (3), the curve has no multiple points. 2. u = Aif - (25 - x 2 )(x 2 + 7) = 0. du dx = 2x{x i + 7) - 2z(25 - x 1 ) - 0. Fig. 34. From these equations we find (a) the z-turns, (5, 0), (— 5, 0), (±•=7,0); (b) the y-turns, (3, ± 8), (0, ± f V7), (- 3, ± 8). DIFFERENTIAL CALCULUS AND PLANE CURVES. 155 The figure (34) is only an approximate representation of the curve. z-turns, (± 4, 18), and (± A ^dyf, ^. y-turns, (0, 4), (0, 0), and :y 5 = 0. = a(2x' > - 3# 2 ) = 0. ± g vw, 19^). w = a: 4 -f- 2ax 2 y du dy du dx ix(x 2 + ay) 0. \ y-turns, (0, 0), (a, — a), (—a, —a); Now there appears to be an #-turn and a y-turn at the point (0, 0), and in a certain sense this is evidently true; but we should regard the result as signifying that (0, 0) is a multiple point of some hind, since x = 0, y = satisfy equations (1), (2), and (3). Let us now determine the character of the point. Dividing (3) by (2), we have dy _ 4# 3 + ^cixy dx Say 2 — 2 ax*' Our object now is to find the value of the slope -~- at the dti multiple point (0, 0). For these values of x and y, -j- assumes the form of -, hence the value required may be obtained by Art. 137. 156 DIFFERENTIAL AND INTEGRAL CALCULUS. We see from Ex. 5, Art. 137, that -f- = and ± V2 at the ClX point (0, 0). Hence the origin (0, 0) is a triple point, the three branches which pass through the point being inclined to the a-axis at the angles 0, tan -1 V% and tan -1 (— 4^2), respectively, as in the figure. See Art. 179. 5. y 2 = «V - x\ z-turns, (0, 0), (a, 0), (— a, 0); y-tnrns, (0, 0), g-Ya/ ±| «'),(- | V% ± i-« J ). The point (0, 0) is a double point of intersection, since at that point dy^ dx ±a. 6. Examine y 2 (ci 2 — x*) — x x = for multiple points. At the point (0, 0), ~~ = ± 0; that is, it has two equal ax values; hence (0, 0) is an osculating point or a cusp; and since the curve is symmetrical with respect to both axes the point is evidently an osculating point of the first species. 7. Determine the general form of the curve y 2 = a 2 x\ When x = oo , y = ± oo ; hence the curve has two infinite branches, one in the first and one in the fourth quadrant. When x is negative, y is imaginary; hence the curve does not extend to the left of the #-axis. When x = 0, y = 0; hence both branches start from the origin. At the point (0, 0), -|-= ± 0; hence, since the curve is symmetrical with respect to the z-axis, the origin is a cusp of the first species. Fig. 36. DIFFERENTIAL CALCULUS AND PLANE CURVES, 157 Again, since -^ = 3a -., the upper branch is convex and Fig. 37. 2Vx the lower concave. 8. Examine the curve (y — ar 2 ) 2 = x h , or y = x* ± x*. Has two infinite branches, one in the first and one in the fourth quadrant, both starting from the origin. For every positive value of x, y has two real values, both of which are posi- tive as long as x < 1, but at the point where x = 1 the lower branch crosses the z-axis. The origin is a cusp of the second species. 178. Tracing Polar Curves. Let f{r, 6) = be the polar equation of the curve. (a) By solving the equation f(a, 6) = for 6, we find the direction of the curve at the point r = a. If a = 0, the values of 6 will be the angles at which the curve cuts the polar axis at the pole. dv {b) By solving the equation -^ = for we find the values of for which r is a maximum or minimum, or the r-turns, at which the curve is perpendicular to the radius vector. 9. Trace the curve r = a sin 3#, Fig. 38. {a) Making r — 0, we have sin 36 = 0; hence 6 = 0, \n, \n, which are the angles at which the curve cuts the polar axis at the pole. civ (b) -jT2 = 3« cos 36 = 0; hence the values d6 TC of 6 at the r-turns are \n i — ,- |-7r, at which Fig. 38. points r ■■ Since dr d6 -a, -{-a, respectively. 3a cos 36, r increases from to a, while 6 increases from to \n\ r decreases from a to — a, while 6 increases from \n to \n\ r increases from — a to -f- a, while 6 increases from \n to f 7t; and r decreases from a to 0, 158 DIFFERENTIAL AND INTEGRAL CALCULUS. while increases from fzr to n. Further revolution of the radius vector would retrace the loops already found. 10. Trace the curve r = a sin 20,. Fig. 39. From this and the previous example, we infer that the locus of r = a sin nd consists of n loops when n is odd, and 2n loops when n is even. 11. Trace the curve r = a cos cos 20, or Fig. 39. r = fl ( 2 cos 3 - cos 0), Fig. 40. 12. Trace the lemniscate r 2 = a? cos 20, Fig. 41. '0 Fig. 41. 179. The character of multiple points in rectangular co- ordinates may often be more easily determined by changing to polar co-ordinates, and applying (a) of Art. 178. Thus, in Ex. 4, Art. 177, make y = r sin and x = r cos 0, divide by r 3 , and we have r cos 4 + 2a cos 2 sin - a sin 3 = 0. Now making r = 0, and we have sin 0=0, and tan 2 = 2; that is, the angles at which the curve cuts the #-axis at the origin are sin -1 0, tan -1 V2, tan -1 — V2. Trace the following curves: 13. The Cissoid, y\2a - x) = s\ 14. The Conchoid, xy = (6 2 - ?/ 2 )(rt + y)\ DIFFERENTIAL CALCULUS AND PLANE CURVES. 159 15. The Witch, (x 2 + ±a 2 )y = 8a\ 16. The Lituus, rV~6 = a. 17. The Parabola, 2 r = a sec - . Z 18. The Curve, • 3# r — flsm -. 19. The Cardioid, r = a(l - cos 6). 20. The Hypocycloid, x* -f y % = a § . 21. Examine a?/ 2 = a; 3 — bx 2 for multiple points. (0, 0) is a conjugate point. 22. Prove that y 2 = x b and (y — x) 2 — x* have cusps of the first species at the origin. 23. Prove that x* — 2ax 2 y — axy 2 -j- a 2 y 2 — has a cusp of the second species at the origin. 24. Prove that x* — ax 2 y — axy 2 -f a 2 y 2 = has a conjugate point at the origin. 25. Prove that the multiple point of ay 2 = (x — a)'\x — b) at (a, 0) is (1) a conjugate point if a < b, (2) a double point if a > b, and (3) a cusp if # = b. CHAPTEE VIII. GENERAL DEPENDENT INTEGRATION. FUNDAMENTAL FORMULAS 180. The differentials in the following twenty-two formulas are the fundamental integrable forms, to one of which we endeavor to reduce every differential that is to be integrated by the dependent process (Art. 51): / v n+l v n dv = — — - + a /dv — = log (v) + C, or log v + log c = log cv. 3. r —£^ = r=:vers- 1 ?;+ (7. V2v -" y = log (v + iV ± m) + t a 22. ^ v* ± m In these formulas v may be the independent variable, or some function of this variable, and the process of integration consists largely in reducing or transforming any given differen- tial into one of the above forms. 162 . DIFFERENTIAL AND INTEGRAL CALCULUS. REDUCTION AND INTEGRATION OF DIFFERENTIALS. 181. Reduction of Differentials is the process of reducing them to integrable forms, and is effected chiefly (1) by constant multipliers, (2) by decomposing or separating them into their integrable parts, and (3) by substitution. 182. Reduction of Differentials by Constant Multipliers. Principle. The value of any differential of the form c I dv remains unchanged if dv be multiplied and c be divided by the same constant. EXAMPLES. Find the following: _ C ndv . n . . v „ 1. y = I — ■ — -. Ans. — - tan" 1 -=. + C. We reduce this to the form of 17, Art. 180; thus n C dv nVa / Wal y — - \ r = / ■ / — r 9 = Answer. /itdv v , (See 15, Art. 180.) Ans. n sin" 1 -j=. + C. Va - v 2 Va r f ' _ n T dv _ nVa f jv_V Va! = Answer. Let the student compare these results with formulas (13) and (15), page 65, and in a similar manner deduce formulas (14) and (16) on that page. GENERAL DEPENDENT INTEGRATION. 163 _ 3 r (2x + 2)dx _ 3 / V(z 2 -f 2a; + 5) V ~ %J x* + 2x + 5~~2J x* + 2x-\-5 - An s. (Formula 2, Art. 180.) ^ */2-9a; 2 V2 r C %dx . _ , _, 6. / - vers -1 3a; + (7. VQx — 9a; 2 Mx 2-f W |/14 """ ~ r 2 r ?>dx 3 x , A /7 8. /V^« i tan- 1 a; 2 +(7. ^ 1 + x* r dx i tan _i ^ + a " t/ l-f-5o; 3 * ^5 P dx 1 , ,- , _. -1. / . :. — sec" 1 aVf + 67. 47 xV2x* - 3 ^3 !. / ._ | vers * -tt+ C. ;. f " ** , * cos" 1 a;^ + (7. ^ V3 - 2a; 2 ^2 1 fx^TQ' itan-^+a 164 DIFFERENTIAL AND INTEGRAL CALCULUS. 15. f ° dx T Vb sec- 1 xV* + a J xVZx* - 5 16. f dx , i^sec- 1 ^ +C. J Vbx* - Zx* Some of the preceding examples may be conveniently solved by formulas (19) and (21), page 65. REDUCTION OF DIFFERENTIALS BY DECOMPOSITION. 183. The process of reducing differentials to integrable forms consists largely in separating them into their integrable parts. 184. Elementary Differentials. In these the necessary reductions are effected by the simple operations of algebra. EXAMPLES. Find the following: 1. y ^tfo s- t log (4z 2 + 1) + I tan" 1 (2x) + 0. r{?>x -f5) _ r 'Sxdx r $dx + 1 «/ 4ar + 1 ' t/ 1 + 4z a 2 . f^J^Ldx. -(l-^^-Ssin-^ + a */ |/l_z 2 3. A ( 2 ~ 5 ^ _ + 2) 3 + C. The first step in the above and similar operations is very simple, and it is our present purpose to show how the second step, the separation and integration of fractions whose denomi- nators contain a higher power of x than the numerator, may be effected ; and to render the process as simple as possible we shall apply it to particular examples in each of the four cases that may occur. GENERAL DEPENDENT INTEGRATION. 169 188. Case I. When the simple factors of the denominator are real and unequal, EXAMPLES. 7 (x + l)dx 1. Integrate dy = ^ 3 + 6 ^ + 8 ^ The roots of z 3 + 6z 5 + 82 = 0, are 0, - 2, and -4; hence the factors of x 3 + 6z 2 + 8z are a;, x + 2, and a + 4. s + 1 -4 5 (7 ' Assume ^ + ^ + 8:g - - + ^2 + ^+4' ' ' (1) Clearing (1) of fractions, we have x + 1 = A(x + 2)(a> + 4) + i?(z)(z + 4) + C(s)(s + 2), (2) or x+l=(A + B+C)x* + (GA + 4B+2C)x+8A. Equating the coefficients of the like powers of x, we have A + B + C=0, 64 + 45+20=1, 84 = 1. Solving these equations, we find A — \ y B — \, and C = — f. Substituting these values in (1), we have x + 1 __1_ 1 3 x 3 + 6z 2 + 8x 8x "*" 4(a + 2) S(x + 4)' dx , dx 3dx 8z~ r 4(z + 2) 8(z+4)' , />c£e . . f* dx » f* dx = i log a + f log (as + 2) - f log (x + 4) + i log e. . l/cx(x + 2) a 170 DIFFERENTIAL AND livTEGRAL CALCULUS. The values of A, B, and C may be obtained from (2), thus Making x = 0, we have 1 = 8 A Making x — — 2, we have — 1 = — 4i? Making x = — 4, we have — 3 = 80 .: B= I Principle. In this case, to every factor of the denominator, as x — a, there corresponds a partial fraction of the form ■ . x a Find the following: 3dx , (x — 2 \* c C Zdx , X — 7£ \* t n J x* — a 2 * T x + a */5^TT log[(*-l)=(* + 2)V]. 5 - Z ^+V-to - log ^— W+8)'+ G - 6 . /*+^J fc £ + %. + 4, + tog^fe^-* + a J x — 4:X 3 2 & (a; + 2) 3 189. Case II. 'TPften some of the simple factors of the de- nominator are real and equal, EXAMPLES. , ' _ , , , (x a + x)dx 1. Integrate dy = (g , 2) , (g . iy Assume -, ^j-, rr = 7 ^ H „ + (x - 2)\x - 1) (x - 2)* * x - 2 * x- 1 Clearing of fractions, we have z s + x = 4 (a - 1) + B(x - 2)(x - 1) + (7(3 - 2) 2 . GENERAL DEPENDENT INTEGRATION. 171 Making x — 2, we have 6 = A; .-. A = 6. Making x = 1, we have 2 = C; .-. C = 2. Making a; = 0, we have = — A -f 2B -f 4C ; .*. 2? = — 1. a 2 + a 6 1 2 •• (x-2)\x-l)~ (x-2f x-2 + ;~T ^~~ (x-2y~x~^2 + x 1' y V (3 _ 2 ) 2 e/ 3 - 2 + V s _ 1 = ~ ^ZT2 ~ lo S (« - 2) + 2 log (x -1) + C = ^(^—27-5^+^ Principle. In this case, to every factor of the form (x — a) n there corresponds a series of n partial fractions of the form A B K (x - a) n ' (x - a)"- 1 ' " ' x — a Find the following: _ P(3x - l)dx . , , ox 8 , n r ^x 1 + 9z - 128)^ d - J (x - 3)\x + 1) • \-17log(x-Z)-8log{x + l) + a 4 -/?+TO+i- i^s + ^C + D + a 6 - J ^+W dx - 2WT7 + 3 los W+l + a 172 DIFFERENTIAL AND INTEGRAL CALCULUS, 190. Case III. Wlien some of the simple factors of the de- nominator are imaginary and unequal. EXAMPLES. 1. Integrate dy = (g + ^*. + 4) . Here the two simple factors of x* -f 4 are x -f- 2 V — 1 and # — 2 4/ — 1 ; we may take these factors and proceed as in Case I, but the integrals obtained would involve the logarithms of imaginaries; to obviate this, we assume x __A_ Bx+ C (z + l)(z a + 4) x + 1 l z 2 + 4 * Clearing of fractions, we have x = A(x 2 + 4) + (Bx+ C)(x + 1). . . . (1) Differentiating (1), we have 1 = 2 Ax + B(x + 1) + Bx + C. . . . . (2) In (1) making x — — 1, we have A — — \. In (1) making x = 0, we have C = — 4^4 = -f f. In (2) making x = 0, we have B = 1 — C — \. Principle. /w £te case, to every factor of the denominator of the form (x — ay -f 6 2 ^ere corresponds a partial fraction of Ax-\- B the form (x _ a y + y GENERAL DEPENDENT INTEGRATION. 173 3 - /ftwt?' 4 tan_1 ■ - * tan_1 1 + a 5 - f^fi d *- , + ilog^- l/3-tan-^ + C. 6 . A«- 1 log ^±^|±_1 + tfj tan- f^ + a. 191. Case IV. Wlien some of the simple factors of the denominator are imaginary and equal. EXAMPLE. 1. Integrate dy = ^ + ^ _ ±y 1 Ax + B , Cx + D , ^ Assume , . , D , 2 , -r- = , , , „ a + - , Q 4- (z 2 + 3) 2 (z - 1) (a; 2 + 3) 2 ' x* + 3 ' x - 1 Clearing of fractions, we have 1 = (Ax + 5)(«-l) + (Cx + Z))(z 2 + 3)(a-l)+.tf(a; 2 +3)". (1) Whence A = -\, B = - ±, C = - T V, 1 »+_!_, 1 (a; + 1) , 1 tfqg «y - - 4 (a .. + 3) .«* 16 z 2 + 3 + 16 x - 1" -^7-in " ^^g (* 2 + 3) - -^--tan- 1 - * 8(z 2 + 3) 32 °^ ' ' 16 ^3 ^3 + ^i og(a; _ 1) _iy > _^_ The integration of differentials of the form -, — _ ' 1 . mt may 5 {ax 2 -\- b) m J be more conveniently obtained by Art. 211 or 215. 174 DIFFERENTIAL AND INTEGRAL CALCULUS. REDUCTION BY SUBSTITUTION. 192. Irrational Differentials. To integrate an irrational differential which is not of one of the known integrable forms, we first rationalize it, and then proceed according to the previ- ous methods. To show, in a simple manner, how rationalization is to be effected, we shall apply the process to a few particular examples. EXAMPLES. Find the following: (2 Vz + i)dx ' V J 2\ / x(x + 3Vx + 5)' Make x = z 1 ; /. dx = 2z dz. y = A%XT+6) = log v + 3z + 5 > = l0 « 1* + 3 ^ + 5 )- 2 A> Vx + l)dx -n ~. tan" 1 Vx + C. 2(x k + x*) T 4. / =. [Make x = z\] sin -1 Vx -f- C. 3 r # a r 1 — x* 6. /%^-v [Makea = « 6 .] 87 «;* + ** 2 1/*- 3fz + 6f^- 61og(l + fe)-fC. /■ (s-l)efo |/^_ 2 GENERAL DEPENDENT INTEGRATION. 175 193. When a -f- to is the only part having a fractional exponent. Assume a -\- hx = z n , where n is the least common multiple of the denominators of all the fractional exponents; then the values of x, dx, and each of the surds, will be rational in terms of z. EXAMPLES. Find 1. y = / x 3 Vl -f xdx. Assume 1 -+- x = z 2 ; then 4 1 + x = z. ... (1) Also x-z*-!, x* = (z°-iy, . (2) dx = 2z dz. . . (3) Multiplying (1), (2) and (3) together, we have fx % VT+xdx =f%z 2 (z 2 — l) 2 dz = ^ - K + \z* + C '-/ xdx 2(x — 2 )4^ 1 +x |/l + z 4. y z(a -f- z) 4 efc. ^(0 + a)*(4z - 3a) + C. 1 94. When y a -f 6a; + x * or ^« + to — # 3 is the only- surd involved. A differential containing no surd except Va -\-to-\- x 1 can be rationalized by assuming Va -\- ~bx-\- x 2 = z — x; and one containing no surd except Va -f- to — x 2 can be rationalized by 176 DIFFERENTIAL AND INTEGRAL CALCULUS. assuming Va -f- bx — x u = (x — r)z, where r is one of the roots of a -j- bx — x 1 = 0. The process is illustrated in integrating the following im- portant differentials (see Ex. 30, 31, page 67). 1. Find y = f- Va + bx + x* Assume Va + bx + x* = z — x; then - a a -j- bx = z* — 2zx, x = 2z + b' 2(z' + bz + a )dz ax ~ (2z + by ' (l > 1 2z + b (2) Va + bx + x' z' + bz + a = log (2« + b) + a .: f- r—^r = log (2* + 6 +2 Va+bx+x') + 0. ^ Va -\-bx -\-x When b = 0, 2. Find y = ^ |/a -}- &z — as 2 Represent the factors of a -j- bx — a; 8 = by # — r and — x, and assume |/# + bx — x* = V(x — r){r' — x) = (x — r)z, GENEBAL DEPENDENT INTEGRATION 177 rz* + r' then t' — x = (x — ?-)z 2 , x z 2 + 1 ' dx - Hr-r')zdz - 1 _ *' + * / 9 v " (s a + l) 2 ' * * {l) Va + te _ ^ " (r' - r)s * ' ^ (1) X (2), f dX = = - 2 I \/a + l)x-x 2 J 1 + 2' - 2 tan- 1 2 -f C. ... ^ ^ = - 2 tan- fA^-* + a When 5 = 0, ?• = + i/a, and r' = — Va, we have 2 tan -1 \ / — — -! + C. Va - x' 1 V V V- a — x 3. */ a V2 + a; - a; 2 ^2 \ ^2 + 2x + V2 - a;/ * Assume V2 + a; - x 2 = V(2 - x)(l + a;) = (2 - z)z; then * = 7TT' ; & "' = ^p and i/2 + x - ¥ =7Ti' r dx _ r 2dz 1 . zV2 — l J x V2 + x -^ ~ J ~z 2 ~ 1 ~~ 1/2 ° g z | 7 2~ + 1 ' 4 " Atf?5^i »t«-M* + •* + &=!) + a /» rgWa; t - 1 (a; + 3) 13 + 2? - x 2 °- y wp^7" 3 sm ~i 2 • 178 DIFFERENTIAL AND INTEGRAL CALCULUS. 195. Binomial Differentials. Differentials of the form r x m {a + bx n )'sdx, where m, n, r, and s represent any positive or negative integers, are called binomial differentials. 196. To determine the conditions under which a binomial differential is integrable. T I. When - is a positive integer the binomial factor can be s developed in a finite number of terms, and the differential exactly T integrated ; and when — is a negative integer the differential is a rational fraction whose integral can be obtained by the method of Art. 187, 212, 214, or 215. II. Assume a -f bx n = z s ; .*. (a -f bx n )s = z r , . . (1) i (z s — a^ x = p-f)> -=(-^r ■ • « and dx = ^rU^)- Multiplying (1), (2) and (3) together, we have m + 1 z w (a + &&")• dx = ^z r + s - 1 (— ^ — J <&. . . (4) The second member of (4), and therefore the first, is in- tegrable when — - is a positive or negative integer, by Case I. III. Assume a + bx n = z s x n ; .\x n = a(z s - b)- 1 , 1 1 m _ m x = a*(z s - b)~~~ n , x m = a»{z s — b) », .... (1) a + fc»" - ~^j, (a + ^ n )* = «'(** ~ &)~ ^dz y ... (2) 1 z s — b GENERAL DEPENDENT INTEGRATION. 179 S - --1 and dx — a*z a -\z 8 — b) » " dz. (3) Multiplying (1), (2) and (3) together, we have r „ m + 1 r / m + 1 r \ z w (« + fe n )^r = a » + *(* s - b)~^ n * h r + s ~ x dz. (4) By Case I, the second member of (4) is integrable when j- — is a positive or negative integer. r Hence, x m (a -f- bx n )*dx can be integrated by rationalization: 7Yh I 1 I. WJien is an integer or 0, by assuming a -\- bx n = z s . XT - _ m 4- 1 , r . . , _ , 11. It ^ew 1 — is an integer or 0, by assuming a -f £>£ n = z s x n . "When the differential reduces to a rational fraction, which is i m + 1 , -. • , • m + 1 r ' ■ „ the case when + 1 1S a negative, or ! 1 f-1 aposi- n n s tive, integer, it is less laborious to integrate by a method to be subsequently given. 1. Find A 5 (l + a 8 )* EXAMPLES. dx. TT m + 1 5 + 1 . Here — ■ = — - — = 3, an integer, and s = 2 ; hence we assume 1 + Z 2 = 2 2 ; .-. (1 +*■)* = *, . . . . (1) x* = z* - 1, z 6 = (^ + l) 3 (2) Differentiating (2), 6x b dx = 6(^ 2 + l) 2 z af».-. ■ (3) 180 DIFFERENTIAL AND INTEGRAL CALCULUS. Multiplying (1) and (3) and dividing by 6, we have /V(l + z')*dx - yV + lyMz = J{z' + 2z> + z')dz = K + K + V+o = {(I + *•)* + 1(1 + *•)* + J(l + «.)•+ ft Find: 2. yva + *•>*€&. (i + x>) i (^~^\ + c. 3. /V(l + z'fdz. A( l + ^)V _ | (1 + j,)! + A(1 + ^i + a 4. yV(l + zY'dz. (1 + ^ , )*(^f : - 2 -) + 0. t>. fz-M + zr'dz. -5S=i + a Vl + uri + r Here m + l_-2 + l__l. m + 1 r__l 3__ ~1^-— ~2 - 2' and n +s~ 2~2~ 2 ' an integer; hence we assume 1 + *■ = zV; .-. (1 + x*)- % = (Z ' 7, 1} ; •.. (1) ■ = <-l; . . . . (2) -77^—. • • (3) OENEBAL DEPENDENT INTEGRATION. 181 Multiplying (1), (2), and (3) together, we have = -{* + -) + c, where z = -Vl 4- x 2 . x 6. / '(1 + af)~*dz. [ro = 0.1 —JL= + a J Vl + x 2 7. fx'il - 2x T ^ - (1 + ^)(1 - 2x^ + c 8. fx'Ha + xT*dx. — + 2a . + a INTEGRATION BY PARTS. 197. Integrating both members of d(uv) = udv + ttf?M and transposing, we have / udv = uv — I vdu, (A) which is the formula for integration by parts. It reduces the integration of udv to that of vdu, and by its application many differentials can be reduced to one of the elementary forms. EXAMPLES. 1. Find / x* log xdx. Assume u — log x; then dv = x 2 dx, du = — -, and v = / x' x or < b - dx /dx — — — -. — ■ can be found. a + b sm x * * m m • a d% ., dx 204. To integrate and . sm x cos x /dx r idx _ Asec 2 \x jdx m sin x ~ J sin \x cos \x J tan \x * I 4^— — log tan \x = log y - •/ sm x . 1 cos a; sin a; & . 1 + cos a; /" dx r dx Agam ' J^ = JMf^) = -log tan (~ _|j-f & GENERAL DEPENDENT INTEGRATION 189 APPROXIMATE INTEGRATION. 205. The number of differentials which can be integrated xactly is comparatively very small, yet the approximate value of the integral of any differential may be found when the dif- ferential can be developed into a convergent infinite series each of whose terms is integrable. This is the last resort in separat- ing a differential into its integrable parts. EXAMPLES. Find the approximate integral of the following: : ., dx x x 2 x 3 x* , , Expanding ——- by division, we have 1 1 x , x 2 x 3 ■ , a 4- x a a 2 a a y = f ( ^- x a^i-¥ +&to ) dx - o j dx By division, 1 + x 2 o j dx 3. ail = — . y = x J VI + x 2 By the Binomial Theorem, 1 y = x - x 3 x h ~ 3~ + 5~ x 7 7 + etc. 1 - - X 2 + X* -x« + etc. x ' + 2.3 ' \ dx 6 2.4. 3. ox 7 + etc. 5 2.4 .6.7 irei = 1 a, _?_! + 3x* 2.4 — etc. Vl + x 2 4. dy = x i (l- x'fdx. y = %x%- \z* - &x* - etc. Develop (1 — x 2 f, multiply by x*dx, etc. 190 DIFFERENTIAL AND INTEGRAL CALCULUS. 5. dy = z 3 (cos x)dx. y = ~ - ^ + ~ - etc. /I ic 2 3# 4 \ 6. dy = x> sin- 1 xdx. y = x 1 ^- + — -f — -f etc. J. 206. Development of Functions by Exact and Approxi- mate Integration. Two or three examples will suffice to illus- trate the process. EXAMPLES. /dx — — = log (a + x), exactly (omitting C); /dx x x 2 x 5 ^ + i = a ~ *? + 3? ~ et °" a PP roximatel y- x x 1 X* .'. log (a + x) = — + --;; — etc. a 2a 2 3a /^ ■ - _ = log (» 4- Vl 4- x 2 ), exactly: Vl+x 2 , r dx x* 3x b . , _ / — = x 4- — etc., approximately. «/ Vl + x 2 6 ' 40 V^ J ,-. log (x + Vl + x 2 ) = x--+~ - etc. 3. / - — : — - = tan -1 x, exactly; t/ J. —r— X dx /ax x ., x x' , . , _ j-— v = a? .— y + -g ^ + etc., approximately. .t 3 £ 6 # 7 .'. tan -1 x ~ x — V + "^ tt + etc. 3 ' 5 CHAPTER IX. INTEGRATION— (Continued). INDEPENDENT INTEGRATION. 207. Increments Deduced from Differentials. We have seen that the increment of a function is the sum of the differ- ential and the acceleration; hence, when the former is known, we can find the differential by simply removing the acceleration. Taylor's formula enables us to reverse this operation in many cases, and find the increment when the differential is known. Let u=J(x). Increasing x by It, we have, by Taylor's formula, u + Au = f(x + h) n Au=f(x + h)-f(x) 2 ' ' " J yuf \n =/'(*)* +/"(*)? + • • • / n w4r • • - ( A ) in which du =f'(x)h. Therefore, when the differential of a function of x is known, the increment may be found by taking the successive derivatives of the differential coefficient, and substituting them in (A). When f n (x) = 0, and each of the subsequent derivatives of f(x) = 0, the series will be finite and express the exact value of Air, otherwise the series will be infinite, and, if convergent, will give the approximate value of Au. 191 192 DIFFERENTIAL AND INTEGRAL CALCULUS. EXAMPLES. 1. If du — (x* — 5x + 6)dx, what is the value of Au? Here /'(se) = a 2 — 5x -f- 6. Differentiating this, we obtain /"(a) = 2x - 5, /"'(a) = 2, / iv (z) = 0. Substituting these values in (A), we have Au = (x 2 - 5x + 6)h + (2a; - 5)^ + ~. 2. If du = (x 2 — 3x — 10)h, what is the value of Au ? Au = (x* - Sx — 10)h + (2x - 3)£ + | 3 . 3. If du = (x 3 — 7x 2 -j- 12x)dx, what is the value of au ? au = (3x* - Ux + 12)^-+ (3a - 7)^ + h * ■ 4. Find aw when du = sin a fc h* . h 3 h* au — cos x - sin x — — cos a; — + etc. 5. If du — (Vl -\-x)dx, by how much will u be increased when x is increased by h ? z/w = (Vl + a;)A+ (1 + x )~^~ (1 + a)"* — + etc. 6. Find aw when *7w = log x dx. h> h 3 h* h> au = ^-W + i^~2-ox-* + etG ' 7. A function is increasing at the rate of 4:X 3 dx; find its suc- ceeding increment. Au = 4x 3 h + 6x*h 2 + ixh 3 + ^ 4 » 8. At the end of ^ seconds the velocity of a body is ds — = (3£ 2 — 2#) ft. per second; find the distance it will travel the following second, dt being the unit of time. As = (3f - 2t)dt + (3* - l)df + aT. 9. The rate of acceleration of the velocity of a body is dv — - == (St -f- 4) ft. per second; INTEGRATION. 193 find the increment (1) of the velocity (v), and (2) of the distance (s) for the following second. Av = (3t + 4)dt + \dt\ v = y (3* + 4)<7tf = ff + 4* + (7. Js = (ff + 4* + 0)<« + (it + 2)tff + \dt\ 208. Increments as Definite Integrals. In Fig. 5, where u = area of OBPA, Au — the area of BCP'P, which is evi- dently the integral of du between the limits x and x -\- h. In general %J X >x+h f'(x)dx=f(x + h)-f(x). For, since df(x) -f'(x)dx, ff'{x)dx =f(x) + G. /*x+h — _ x+h I f'(x)dx = \f(x) + C\ =f(x + h) -f(x). *J x * Therefore (A) may be written r x+h h* h 3 h n J f'{x)dx =f'(x)h+f"(x)^ + f'"{x)j . . ./%e)j-...(B) By this formula we can obtain exactly, or in the form of an infinite series, the definite integral of any function of a single variable, and the operation does not involve the reversing of any of the formulas for differentiating. But, in general, this method is much inferior to that of dependent integration, since by the latter many differentials can be integrated in finite terms which by the former could be expressed only in the form of an infinite series. However, it forms an important part of the theory of differentials and integrals, and is often useful as a method of approximation. More convenient formulas for practical purposes will be derived from (A), but before doing so let us apply (B) to the following illustrative examples. 194 DIFFERENTIAL AND INTEGRAL CALCULUS. 1. Find the area BCP'P, the equation of APP f being A y = x 2 — lx + 12, where x = OB, y = BP, and BC = h or dx. Let w = area of OB PA, then dw = ydx. y=f'(x)=x 2 -7x + 12, f"(x) = 2x-7, f'"(x) = 2, /*(*) = 0. Substituting in (B), we have ' x+h h* h 5 (y)dx = (x* - Ix + 12)^ + {2x - 7)y + y = area of BCP'P. I Cok. I. To find the area of OEA we make x — and h = OE — 3, and obtain 13J. To find the area of EnF, we make x = 0^ = 3 and ft = j£F = 1, and get — f Let the student solve the following in a similar manner. 2. The equation of a curve is y = x* — '6a?* -f- Ha? — 6; find the areas of the two sections enclosed by the curve and the axis of x. i; — i. 209. A More Convenient Series. In (A), by makiug x = 0, then making h — x, and writing / f'{x)dx for f{x), we have //'<*)** =/(0) +f'(0)x +f"(0)~+ . . -f"(0)y . . (0) This formula may be obtained by developing f'{x) by Maclaurin's formula, multiplying by dx, and integrating each INTEGRATION. 195 term separately, but as we are now exemplifying the method of independent integration, we will apply (C) directly to one or two examples. 1. Find f X {Zx* - 14a + 5)dx. Here f'{x) = 3z 8 - 14a; + 5, .-. /'(0) = 5; f"(x) = 6x- 14, .-. f"(0) = - 14; f'"(x) = 6, /./'"(0) = 6. Substituting in (C), we have J^hx* - Ux + b)dx = x 3 - W + 5x. 2, J X {x % - 6x 2 + !)dx. ± x * _ 2x 3 + 7a?. 3. f*(3z A - 2x* + x*)dx. |^ _ ^ + i^. 210. Bernouilli's Series. In formula (B), by making /x nx f'(x)dx = — I f'(x)dx, we have £ f'(x)dx =f'(x)x -f"(x)^- +...-(- l)"f"Wy • (D) This formula, called Bernouilli^s Series, like formulas (B) and (C), shows the possibility of expressing the integral of every function of a single variable, in terms of that variable, since the successive derivatives /"(#), /'"(#), etc., can always be deduced from f'{x). Hence, in all cases where the series are finite or infinite and convergent the integral may be ob- tained exactly or approximately. In finding ff'{x)dx by (B), (0), or (D) the limits of the difference between the approximate value found and true value may be determined as in A 6 of the Appendix. 196 DIFFERENTIAL AND INTEGRAL CALCULUS. INTEGRATION BY INDETERMINATE COEFFICIENTS. 211. The process of integrating binomial and trigonometric differentials by successive reductions is generally very tedious, and it is our purpose now to present a method which is gener- ally less laborious, and which is also applicable to many other classes of differentials. Let u r vdx be the differential to be integrated, where u and v are functions of x, and let us assume CvTvdx — u r+1 f(x) + kfu s v x dx, . . . . (E) in which it is required to find the function f(x) and the con- stant k. In examples where / u r vdx can be expressed under the form of u r+1 f(x), we shall find k = 0; and when this is not the case, / u r vdx will be determined in terms of / u s v^dx, which we can generally make of a more elementary character by as- signing suitable values to s and v % . Another advantage of expressing the required integral under the form of (E) arises from the fact that u r+1 /(x) often vanishes for the desired limits of integration, in which case the definite integral depends on the last term only. Differentiating (E) and dividing by u r dx, we have v = (r + l)fjjM+ «/*(*) + *»,«•-.. . . (F) The simplest and easiest method of solving this equation for f(x) and k is by indeterminate coefficients, as illustrated in the following examples. 212. Case I. When k = 0; — Independent Integration. EXAMPLES. 1. Find y (1 + x")~ k x\lx. INTEGRATION. 197 Comparing this with (E) we have u — 1 -f- x*, r = — -J, j— = 2x, and v = x\ Substituting in (F), we have ax * = &z) + (l + aW{*) + && + *)"*- . ■ (1) The first member of this equation being of the fifth degree the second must be also; hence, f(x) must be of the fourth de- gree, and since u involves only the second power of x, we may assume f(x) = Ax* + Bx 2 + (7, .\ f\x) = ±Ax 5 + 2Bx. Substituting in (1), and arranging in reference to x, we have x* = bAx b + (3£ + ±A)x* + (C + 2B)x + kv^l + x*) s+ K 1 = 5A, 3£ = - 4.4, C = - 2B, Jc = 0, or A = l B = -&, C=&. These values determine f(x), which substituted in (E) gives f(l + xY'v'd* = (1 + * ! )*[-K - ^ + &*] + ft 2. Find yvu + *•/**. - (i + «■>• (i.. - r y + a Here u = 1 + a", r = -J, -7— = 2z, and «; = a" 6 ; .•. (F) gives *-* = ft^(a) + (I + «V(*) + *» I (l + ^)- 1 : . . (1) In order that the two members may be of the same degree we assume f(x) = Ax' 5 + Bx~ 3 + Cx- 1 ; .-. f'(x) = - 5Ax-«-3Bx- 4 - Cx~\ Substituting in (1), and arranging in reference to x, we find A = — \ 9 B = T \, G — 0, h = 0; this determines f(x); which substituted in (E) gives the desired result. 198 DIFFERENTIAL AND INTEGRAL CALCULUS. A careful inspection of the previous examples suggests the following rule for determining the form of f{x) for binomial differentials of the form (a -j- bx n ) p x m dx, v being x m : I. When m is positive the highest exponent of ' x infix) will be m — n -f- 1. II. When m is negative the algebraically lowest exponent of x in fix) will be -$ m -\- 1. III. The remaining exponents decrease or increase alge- braically by n. The rule is also applicable when u is a polynomial in which n is the highest exponent of x, provided that the exponents of x in f(x) increase or decrease by the least difference between the exponents of x in u. When r is a fraction and u a polynomial of a higher degree than the second, the differential cannot ordinarily be inte- grated; or, more accurately, its integral cannot ordinarily be finitely expressed in terms of the functions with which we are familiar. The exceptional or integrable cases are, in general, where u, v, and r are such that it is possible forf(x) to have as many coefficients as there will be independent equations between the coefficients in equation (F), and where k I v x u s dx is or one of the integrable forms. In a differential of any given form the conditions of integrability may often be determined by the present method. r dx Vx 2 + Qx -f 15 f 1 1_ 1_\ _! q Vx* + 6x + 15 ( 1 1 -45 ^ 3 2x* 15, r = — i, and v = x-'- Vx % + 6x + 15 Here u — x 2 -j- § x + 15, r = — i, and v = x~^\ hence the lowest exponent of x in fix) will be — 4 + 1 = — 3, and the others will increase by 1, giving fx = Ax~ 3 -J- Bx~ 2 -f- Ox~ x -\- D. The process can also be applied to many differentials in which v is a polynomial, as in the next example. r 3z 2 -f bx -f 5 , : /q~ i I \ . INTEGRATION. 199 Here u = x* + 2x -f 3, r = — i, y = 3a; 2 -|- 5# -f- 5, and. /(a;) is of the form Ax -j- B. . r ax ^_ri^,i a . + C. Here # = 1 -f- x*, r = — %, v = 1; and making v 1 = 1, s = - h (F) gives 1 = - 53/0*) + (1 + x*)f'\x) + k(l + z 2 ) 3 , where /(a?) is evidently of the form Ax 5 -f ito 3 + (7a;. 213. We have seen (Art. 185) that sm m x cos n x dx can be easily reduced to an integrable form when either m or n, or both, are positive odd integers, or when m -f- n is an even integer and negative. In such cases, m and n being integers, the inte- gration may be effected by the independent method, as in the two following examples; but this method of integrating such differentials is introduced and recommended chiefly for its bear- ing on the cases in which the above conditions do not exist, and which are usually solved by successive reductions. C • s 4 7 5 !~ s i n4 x , 4 sin 2 x , 8 ~| , „ 6. / sin x cos x dx. — cos x — —r- 4- — — — -f — — -f- C. U |_ 9 63 oloj We may make u = cos x, r = 4; then -r~ — — sin x and v — sin 5 x. Substituting in (F), we have . sin 5 x = — 5 sin xf(x) -j- cos xf'{x) -{- Tcv x cos 3 " 4 x, where f(x) is evidently of the form A sin 4 x 4- B sin 2 x -\- C, and hence /'(a;) = (4J. sin 3 x + 2i? sin a;) cos x. .*. sin 5 a; — — 5 J. sin 5 sc — 5B sin 3 a; — 5C sin a: -f- (4J. sin 3 x -f 2i? sin x) cos 2 a; -f- etc. 'Now, substituting 1 — sin 2 x for cos 2 a:, reducing, and arrang- ing with respect to sin x, we have 200 DIFFERENTIAL AND INTEGRAL CALCULUS. (1 + 9 A) sin 5 x + (IB - ±A) sin 3 x + (5 - 2B) sin x -f Tcv x cos s_i x = 0. .-. 4=--i, B=-^ #=-1** * = 0. C dx sin a; f 8 . . 4 . . . ., H . ~ 7. / — s— . — r— — sm 4 a; — - sin 5 x-\-l \ + C. J cos 6 x cos 5 # [_ 15 3 J Make w = cos a\ ?• = — 6, -=— = — sin #, v = 1, 5 = 0, ax v l = 1, and we have from (F) 1 = 5 sin xf(x) -j- cos %f'(x) + h cos 6 a?. Assume f(x) = ^4 sin 5 # -f ■# si* 1 * ^ + Csin x. .'. f'(x) = bA sin 4 x cos # -f- 3i? sin 2 a cos a; + C cos a. Substitute, reduce, etc., and we find A = ~-, B=-~, G = 1, k = 0. This method of integration can often be applied to other classes of differentials, as in the next two examples. /x h ( 2 9 ■ \ x* log 2 x dx. - (tog 2 x --logx + g 5 J + and (making 5=0, v 1 = ^- J * = a /^# sin a; . 5 . 1 sin x cos 7 a; ~~ cos 7 x 6 cos 8 x 6 . 4 cos 4 a; 5.3.1 sin a: .5.3.1 r_dx_ J cos a?* 6.4.2 cos 2 a; 6 . 4 . 2 «/ cos a;* To integrate the last term see Art. 204, page 188. 7. / sin 4 x cos 4 a; fe . „ f cos 6 a; , cos 3 a? 3 cos x~\ 3 , . „ gm .^___ + __ + _-J__( Bn *.oo B *-*)+a Make w = sin ar, r = 2, then v = sin 2 a; cos 4 x = cos 4 a;— cos 6 a:; also make 5 = 2 and v x = 1, then cos 4 x — cos 6 x = 3 cos a;/(a;) + sin xf'(x) -\-h. . . (1) Make /(a;) = ^4 cos 5 a? + J5 cos 3 x-\- cos ar, find f'(x), sub- stitute in (1), for sin 2 x write 1 — cos 2 x, etc., and we find A = -i, 5 = T V, tf= A, and £ = ^. / 7 . a r cos 5 x , cos 3 a: 3 cos x~] sm 4 a; cos 4 xdx = sin a; — -\ — — | — — To integrate the last term, see ex. 11, Art. 185. 206 DIFFERENTIAL AND INTEGRAL CALCULUS. x 2 dx 8. Va? [- X 2 Va '-^ + Y^- lX aJ = ^ dx Va 2 - [ Va 2 - x 2 , 1 _ • a -L log (- 1) 2a' 10 2a 2 x 2 2d 6 to a 4. |/ a * _ ^ . f Va 2 + a; 2 r =p, -f- = nbx 11 - 1 , v = x m , s — r, and v 1 = x m ~ n . Substituting in (F), we have x m = (p + Vjnbx'^fix) + (a + bx n )f'(x) + Jcx m ~ n , (1) where /(#) = Ax m - n+1 , and /'(#) = (m — w -f l)Ax m ~ n . Substituting these values in (1), arranging with reference to x, we find A = L and ]c = < m ~ n + x ) , 5(wp + w + 1)' ' &(wp + Wi + 1)' Ci - ^ n\» «,j x m - n+1 (a + bx n y +1 ,\ / (a + bx n ) p x m dx = —n — \- ■ — — fr— e/ v ' ' b(np + m + 1) By a repetition of this formula m may be diminished by any integral multiple of n. II. Required J (a -\- bx n ) p x m dx in terms of /(■+ bx n ) p - l x m dx. Make w = a -f £# n , r =j? — 1, s =p — I, v = (a + bx n )x m , and v t = 2 W . INTEGRATION-. 209 Substituting in (F) and proceeding as before, we get f(a + bx")Vdx = xm+1{a + *?}* J v J np + m + 1 °L n P C(a _|_ lx«)*-H m dx. (B) Each application of this formula diminishes the exponent of a -f bx n by unity. When m or p is negative, we need formulas for increasing instead of diminishing them; hence the following : III. Required / (a -j- bx n ) p x m dx in terms of J (a + fo n ) V+ TC d£. Solving (A) for / (a + bx n ) p x m - n dx, and substituting ?ti -f- w for i», we get t/ v ' a(m -f 1) - »(«*> + » + "* + !) A,, + fc^-^fa (C) IV. Required I (a -\- bx n ) p x m dx in terms of J (a + fa n )* +1 a w rfa;. Solving (B) for J {a -j- &c B ) 3,-1 a: m (fc, and substituting i? + 1 forjj, we find / (a + bx n ) p x m dx = v ,\ ' — «/ 7 aw(p + 1) ^ + ^+m+i /• j^^ (D) 210 DIFFERENTIAL AND INTEGRAL CALCULUS. 216. The approximate integral of many differentials may be conveniently obtained by the method of indeterminate coeffi- cients. The following important example will serve to illustrate the process. Integrate the Elliptic Differential dx Comparing this with (E), we may make u = a? — x*, whence r — — i, y- = — 2x, and v — (a 2 — eV)*, which, when devel- (XX oped by the Binomial Theorem, gives e* \ l_ \ e° , V ~ a 2a X 8a* X 16tf X ' ' ' Substituting in (F), Art. 211, making s = — -J, v 1 = 1, we have where f(x) is evidently of the form fx = . . . Ax b -f- Bx 3 -\- Cx. Proceeding as in Art. 212, we find •■•[•I- <••-»'[• ■«£)'+ (w+s)S" + W + 64+4 AS +T-4" 64 ~ 256 " • "J Sm a" CHAPTER X. INTEGRATION AS A SUMMATION OF ELEMENTS. ELEMENTS OF FUNCTIONS. 217. Hitherto nothing has been said about the magnitude of differentials. Whether they are large or small does not affect the principles which have been deduced; hence we may regard them as small as we please. They are variables whose limits are zero. 218. In the present chapter increments are called and treated as Elements.* Thus Ay or Af(x) (— m J l + m Ji 2 , Art. 24) is an element of the function y — f(x). For convenience the element Af{x) will often be represented by E xy and the differential dy or f'(x)dx by D x , which may be called, respectively, the xth element and the xth. differential of the function f(x). Since D x varies as dx and approaches E x indefinitely as dx approaches 0, D x is called the differential value of E x with respect to dx. The expression ^ 2 [E X ~\ represents the sum of all the ele- ments like E x , or the sum of the successive values of E x , be- tween the z-limits a?, and x r That is, supposing the increment of x to be always li, 2 x [Ex\ = E Xl -\- E Xl + h + E Xl + 2h + • • • E x ^ _ h (or Xl + ( n _ 1)70. 219. A Definite Integral Regarded as a Sum. The prac- tical importance of integration consists chiefly in regarding it * Because sum and element are correlative terras. 211 212 DIFFERENTIAL AND INTEGRAL CALCULUS. as the summation of a certain series. For example, in seeking the area of a curve, we conceive it divided into an indefinite number of suitable elementary areas, of which we seek to deter- mine the sum by a process of integration. The solution of this fundamental problem is effected by the following formula and its corollaries. Suppose that in any function of x, as/(#), we change x from x x to # 3 by giving to x successive increments. The whole change in the value of f{x), viz.,/(# 2 ) — /(#,), must be the sum of the partial changes produced by the increments given to x. That is, or (Art. 208) f**D x = ^\E X ] (A) Formula (A) is not only an expression of the simple fact that the whole is equal to the sum of its parts or elements, but it signifies that the integral of the differential value of an element, between certain limits, is the sum of the successive values of that element, between the same limits. Fig. 43. As an illustration of (A) let us consider the area of BGQF y where OE — x 1 and OG = x r Suppose y=f'(x) to be the equation of AFQ, where x = OB and y = BP. INTEGRATION AS A SUMMATION OF ELEMENTS. 213 Let u = the area of OB PA, and let BC ( = h) be an element of x, then BCDP - du =f'(x)dx = D x , and BCP'P = D x + au = &. (a) Evidently, EGQF = f X *B x , Art. 208. . . . (1) (#) Divide i?6r (= x 2 — aj into w parts, each equal to li, and draw the ordinates a 1 d 1 , a 9 d 2 , a 3 d 3J etc., then EGQF=Ed 1 + a 1 d u + a,d 3 +, .... (2) in which Ed l , a x d 9 , # 2 d 3 , etc., are the successive values of E Xi as a increases by h from a^ to x 9 . That is, Ed x —E Xx , afi^—E^ + hy a 2 d 3 = E Xl +2h> etc. Hence (2) may be written EGQF=^[E X -] (3) Now, equating (1) and (3), and we obtain formula (A). In further illustration of formula (A), let us show that the signification which it expresses is true of / 3x*dx. tJX\ (a) r x *$x t dx = \x l + Cf* = xf-x l \ ... (4) t/Xi X\ (b) Since /'(a) = 3# 2 , E x = 3x*h + Zxh* + h% Art. 207, E Xi =3x*h + 3x 1 h 2 + h 3 ; E Xl + h = 3x x *h + 9xjf -f W; E Xl + 2h = Sxfh + lbxji* + 19& 3 ; E Xl + (n -i)fc= 3?^ + 3^(2^ - 1)A" + (3^ 2 -- 3» + 1)£ 3 . Taking the sum, remembering that x x + nil = x 9 , and, by Algebra, 3 + 9 + 15 + . . . 3(2rc - 1) = 3rc 2 , and 1 + 7 + 19 + . . . (3w a — 3n -f 1) — n a , we have ^[^J = 3nhx* + 30^)^, + (wA) 1 V = (a;, + fiA) 1 - a:, 3 = z 2 3 - x*. ... (5) 214 DIFFERENTIAL AND INTEGRAL CALCULUS Comparing (4) and (5), we see that the results of the opera- tions indicated in (A), when applied to / Sx^dx, are the same. 220. Formula (A) is also true when x 2 — x x (— EG) is not divided into equal parts. Let us suppose x 2 — x x to be divided into the following equal or unequal positive parts : a — x lf b — a, c — b,.. . . I — h, x 2 — I, the sum of which is evidently x 2 — x } ; then we have identically / y dx — I y dx -f- if dx-\- y dx -\- . . jydx > (1) t/ari vxi *Ja *" Jb Ok in which a — x l9 b — a, c — b, etc., may be considered the suc- cessive values of Ax and / y dx, I ydx, etc., the correspond- ing successive values of E x . Hence (1) is the general significa- tion of (A), which the student may easily illustrate with a figure. 221. A Definite Integral Regarded as the Limit of a Sum. In Fig. 43 let us suppose n to increase and h to decrease, nil being always equal to EG. Since the limit of E x -±- D x , as h approaches 0, is unity, the sum of all the rectangles like D x approaches indefinitely the constant sum of all the elements like E x . Therefore limit ^*F-n 1 ^Trrl Substituting in (A), Art. 219, we have i>="S That is, the definite integral / D x is equal to the limit of the sum of all the successive values of D x , as x increases by h from x, to x„. INTEGRATION AS A SUMMATION OF ELEMENTS. 215 For example, let us find the value of / 3x 2 dx. Since D x = Sx^h, we have D Xl = Zx?h, D Xl + h = Zv*h + 6*^" + U\ D Xi + 2h = 3x^h + 12xJ? + 12h% D Xl + {n _ 1)h = Zx?h + Q(n - l)xfi' + 3(w - 1)'*\ Taking the sum, remembering that + 6 + 12 + . . . 6(n - 1) = 3(?z 2 - ri), and + 3 + 12 +» + . ■ . 3(n - 1)« = (W " 1} ^ " X \ we have ^>,] = Sate.! + 8(»- - »)*,*• + (n ~ 1){n) 2 (2n ~ 1] h> = ${iih)x? + 3(^)X - 3(wA)^ + {nhy - \{nlifh +- (^V. Now, making nh = x^ — x 1 , and then passing to the limit by making h = 0, we have j^ 2 x 'P*l = 3 ^ - *>.' + 3 ^ - *.)"*. + fo - *,) 3 which is evidently equal to / Safdx. 222. It is important to observe that, whether an integral be regarded as a sum, or the limit of a sum, integrating is equiv- alent to two distinct operations: (a) If a st«m, as in Art. 219, integrating/' (a:) $» is equivalent to (1) increasing the differential f'(x)dx by the acceleration au to obtain the element E x , and (2) finding the sum of the succes- sive values of E~ 216 DIFFERENTIAL AND INTEGRAL CALCULUS. (b) If the limit of a sum, as in Art. 221, integrating/ '{x)dx is equivalent to (1) finding the sum of the successive values of f'{x)h, and (2) taking the limit of the sum, as h approaches 0. In case (a) all the quantities involved are finite; but in case (5) the limit of each part is 0, and the limit of the number of parts is qo . Both methods have their advantages, and hence both will be employed, more or less, in the applications which follow. Just here the student may profitably read Art. 238, which offers a simple illustration of the significations and practical im- portance of formulas (A) and (B). APPLICATIONS TO GEOMETRY.* 223. Lengths of Curves.— I. Bectangular Co-ordinates. To find the length (s) of the arc APQ between the limits OB = x x and OG = x v (Fig. 44.) d 5 Fig. 44. Here E x = Pd x and D x = Ft = /l -f (^f) V Art. 33, and * The previous applications of Calculus to Geometry, Arts. 62, 64, 65, 66, were limited to the most elementary rules for integration ; in this chapter it is our purpose to extend these applications by the more advanced methods of integration with which the student is now familiar, and in doing so to impress upou him the important principle of integration as a summa- tion. INTEGRATION AS A SUMMATION OF ELEMENTS. 217 #2 limit - r 2 P%i Pd, + d,d, + a A + etc. = 2 Xi W\ or ^^[2).]=^ Z>*. - -XViTS)^ or £V^(§)W. ( C ) EXAMPLES. £ 3 1 1. Find the length of the arc of the curve y = — + — be- tween the limits x, = 1 and x a = 2. dx dy _ x* - 1 / "da 2*? + ^ "2- ~ L 2» 4 ~6Ji 2. Rectify the parabola y* = Aax, using the formula dy y \dyH dy~2a' •■dy- y 4a' + 1 -2a iy +ia - = y V4g + 7 +gb8 / |>+Vg +g\ > Art. 214, Ex. 10. 3. Bectify the curve y = log (x + |/x 2 — 1). 4. Rectify the ellipse ?y 2 = (1 - e 2 )(a 2 - x 2 ). dy ,„ ..a; xVl-e 2 -j^- ■= — (1 — e 1 )- = Art - 97 '> hence Pd, + d x d % + d 2 d 3 + etc. = 2*[Ee] limit A n1 /"'mi - -=r(-+s)>-/X i+ ^') v - (D) 11. Rectify the spiral of Archimedes, r — aO. Heref = 1; ,. s = /*(l + 3 W * /> + O** dr a e/o V ' «v « t/o v ' y r(« 2 + r 2 )* , a . r + ^ 2\ eC ~ e e h 20. Find the length of a quadrant of the curve /*y , toy. a a +«ft + y [a J ^Kb) - U . « + 6 * 225. To find the equation of a curve when its length is given. 1. Find the equation of a curve whose length is x 2 s = |logz+-. ds dx 1 + ^. . i/(d s Y l — x* _dy 2x ~ ; •"' * \dx) ~ ~ 2x ~~dx' 1 — x* Hence, dy = — - — dx, and y = i log z — \x* -f- C EXAMPLES. Find the equations of the curves whose lengths are: *•* = -■& + £• y = s ~^ + C - See Note, p. 219. 222 DIFFERENTIAL AND INTEGRAL CALCULUS. 'x-Y 3;.* = *+-l0 gVr + ]y 4. s = i log (tan #) 1 y — \ log (sin 2#) -j- C. 5. 5 i + 2(n- l)^- 1 * 2(» + l) y = s- n-\-\ + 0. 226. Areas of Curves.— I. Rectangular Co-ordinates. To find the area of the surface between a given curve, the axis of x and two ordinates whose abscissas are x l and x t , we have, Art. 219, «■=./ ydx = 2 [E x ]. . . ; . . (E) For a definite area between the curve and axis of y, we have u= xdy = 2JEy\ (F) ' EXAMPLES. x* 1 Find the area of the curve y = —— -j- .—— between the limits x t = l and # 2 = 2. I " f x* i \ r~# 6 i n 2 51 fo = i lie- + §#r* = Lio- - & J, = 80- 2. Find the area of the circle y 2 = a 2 — x 2 . Area of OB PA = fydx = / (a 2 — x 2 Ydx = — — - — *- -f — sin -1 - t/o 2 2 rt (Ex. 17, Art. 96) X B C Fig. 47. _ (0B)(BP) (0P)(slycAP) _ + = area 0£P 4- area OP A. INTEGRATION AS A SUMMATION OF ELEMENTS. 223 To find the area of the quadrant OCA we have f\a 2 - tffdx - ±na\ I/O The value of n is given in Art. 129. 3. Find the area of an ellipse, a 2 y 2 = cTb 2 — b 2 x 2 . y = - l/V - x 2 ; .'. u=- fv'a^^x 2 dx. i (the area) = -*- / Va 2 — x 2 dx = \nab\ entire area = nab. 4. Find the area of the hyperbola, a*y* = b 2 x 2 — a 2 b 2 . y = - Vx 2 — a 2 : ,\ u = - / \/x 2 — a 2 dx. a a J or bx{x 2 - a 2 ) k ab . , , A/ - 2 5 x , n u = -A_^_ — log (x + Vx 2 - a 2 ) + 0. To find C, we know that when x — a, u = 0; hence 0=- a poga+C; .: C = ~loga. Substituting this value of C, and making Vx 2 — a 2 — -~- f we have Area of hyperbola = -^ - ~ log (^ + 1 J. 5. Find the area of the surface between the arc of the pa- rabola y 2 = ±.ax and the axis of y, \xy. 224 DIFFERENTIAL AND INTEGRAL CALCULUS. 2*21. It is often convenient and suggestive to regard a defi- nite integral like f Z 'y(dx) = EGQF (Fig. 43) as signifying that "the ordinate y (or generatrix PB), moving perpen- dicularly to the axis of x from x=x x to x = x 9 , generates EGQF. Thus, let it be required to find the area of the surface between the parabola y 2 = 4ax and the straight Fig. 48. line y = x. We at once have OP DC = / (PC)dx = / (V *J 1/0 4ax — x)dx This method can be employed, with equal facility, in finding the volumes of many solids, in which case the generatrix is a surface. 6. Find the area of one branch of the cycloid. dx _yfy . . u= _ r tfdy V2ry - f~' J V2ry-y* For the area of one branch we have = %7ir\ (Ex. 13, Art. 214) /*Zr J n V2rv- -y-y which is three times that of the generating circle. '. Find the area of the curve xy a. M a log 8. Find the area of the curve x 5 y — x + 1 = between the .^-limits 1 and 2. £. 9. Find the area of both loops of a*y* — aWx* — b*x\ %ab. INTEGRATION AS A SUMMATION OF ELEMENTS. 22b 10. Find the area of both loops of the curve a e y 2 = a V — x*. 11. Prove that the area of the curve a^y 3 = (a 3 — x 3 )x\ be- tween the ^-limits and a, is the same for all values of a. 228. II. Polar Co-ordinates. In Fig. 46 the area of POD (— ±r*dd, Art. 35) is the differential value of the element POd/, therefore jf V [•te+S-y+'^lr" 7 *^- 2. Find the area of the surface of a prolate spheroid, the generating curve being the ellipse a 2 y 2 = b 2 (a 2 — y 2 ). y = — Va 2 — x*. and ds = \ — 5 ~dx. 9 a r a 2 — x 2 .'. Area = 2 / 2;ry^s = 47T- / (a 2 — e'x*)*dx Jo a Jo . b [~ x(a* - e'x 2 )* , a 2 . . ex~] a — 4^_ _v /, ^_ sm -i _ a L 2 2e a J = 2tt6 2 H sm _1 e. e 3. Find the area of the surface generated by the revolution of the cycloid about its base. /*2r /*2r Area = 21 2nyds = 47T 4/2 r / */o Jo o V2r -y 228 DIFFERENTIAL AND INTEGRAL CALCULUS. it y2r£- f(4r + y)(2r - y?~j 64 ■ = T /Tr ' 4. Find the surface of the paraboloid between the limits x = and z = a, the generating curve being ^ 2 = 4o& i(V8- l)87ta\ 5. Find the surface generated when the cycloid revolves about the tangent at its vertex. - 3 ¥ 2 -arr a . 6. Find the surface generated by the revolution about the axis of x of the portion of the curve y = e x , which is on the left of the axis of y. n {y% _j_ i og (i + 4/2)). 7. Find the entire surface of the oblate spheroid produced by the revolution of the ellipse a?\f -(- #V = a 2 b 2 about its minor axis. ' b 2 1 + e 2^-rt -f n— log; — - 1 — . 1 e 5 1 - e 8. Find the surface generated when the cycloid revolves about its axis. 87rr 2 (7r — J). 9. Find the surface generated by revolving the arc of the cardioid r = a(l — cos 6) about the initial axis. 8 = Infyds = 2?tfr sin 8 Vdr* + rW. ¥*«*• 10. Find the surface generated by the revolution of a loop of the lemniscate r 2 = a 2 sin 26 about the polar axis. 2t« 2 . 230. Volumes of Solids of Revolution. In Art. 32 the volume of the cylinder {ny^dx) generated by the revolution of the rectangle BCDP is the differential value of the xt\\ element of volume; therefore v = 2%[M«] or J™" 2 ?**"*] = *£?*' d *- (J) INTEGRATION AS A SUMMATION OF ELEMENTS. 229 EXAMPLES. 1. Find the volume of a paraboloid, the generating curve be- ing the parabola y 2 = kax. v = n j y 2 dx = 4:7ta I xdx = 2nax 2 . When the curve is revolved about the axis of y, we evidently have v — n j x 2 dy. 2. Find the volume generated by revolving the surface be- tween the parabola y = + Vkax and the axis of y about that axis. [vf o = nfx'dy = nf^dy = £jL = ^x'y. That is, the entire volume is one fifth of the volume of the circumscribing cylinder; therefore the volume generated by the surface of the parabola in revolving it about the axis of y is four fifths of that of the circumscribing cylinder. 3. Find the volume of the solid generated by the revolution of the cycloid about its base. dx = — -/ • ; .*. dv = ny 2 dx = V2ry — y 2 V2ry — y 2 To obtain half of the volume, we must integrate between the limits y = and y = 2r. Jo V2ri J - f That is, v = f of the circumscribed cylinder. 4. Find the volume generated by revolving the ellipse AA f about the tangent X'Xas an axis. (Fig. 49.) Let OA=a, OY=l, O'B = x, BP = y, and BP' = if \ than y = -(a -j- \/a 2 — x 2 ), and y' — -(a — Vet 2 — x 2 ),. 230 DIFFERENTIAL AND INTEGRAL CALCULUS. Supposing P'n = dx, the volume generated by P'm, viz., n(y i — y n )dx or 47r- | 7 a 2 — x*dx> -x- y^ X m / A \ ° J x^ p' y 4 n ^^^^__ O' Fig. 49. is the differential value of the volume of the element generated by P'cdP, with respect to dx. f ±n- Va? - xhlx = 2ttW, (Ex. 17, p. 56) which is the entire volume, being the sum of the volumes gen- erated by revolving all the elements like P'cdP between x— —a and x — a, or the limit of the sum of the volumes generated by all the rectangles like P'nmP. 5. Find the volume of the closed portion of the solid gener- ated by the revolution of the curve («/ a — by = « 3 ^ around the axis of y. 256 nV 315 « 6 * 6. Find the volume generated by revolving the curve (x — 4a)y 2 — ax(x — 3a) about the axis of x, between the ^-limits and 3a. ^-(15 -16 log 2). 7. Find the volume generated by revolving the cycloid round the tangent at the vertex. n' 2 r 3 . 8. Find the volume and surface of the torus generated by revolving the circle x' 2 -f- (y — bf = a 2 about the axis of x. Zn-cvb and An^ab. INTEGRATION AS A SUMMATION OF ELEMENTS. 231 9. Find the entire volume and surface generated by revolv- ing the hypocycloid x* + y* = c? about the axis of X. 32tt0 3 _ 127m 2 and 105 5 10. Find the volume generated by the curve xy* = 4a 2 (2a— x) revolving about its asymptote. <±/r 2 # 3 . x 11. One branch of the sinusoid ?/ = I sin — is revolved about the axis of x; find the volume generated. %7r-ab'\ SUCCESSIVE INTEGRATION. 231. A Double Integral is the indicated result of reversing cVu the operations represented by 7 , » Thus, if -j— -y- — xy*. then u — I I xy"dy dx, which indicates two successive integrations, the first with refer- ence to x, regarding y as constant, and the second with reference to y, regarding x as constant. Thus, ffxtfdy ** =f[j + c )y* d y where C and C x are the constants of integration. 232. Definite Double Integrals. Here both the integra- tions are between given limits. Pc Pa For example, / J x*y*dx dy. This notation indicates that the integrations are to be taken in the following order : 232 DIFFERENTIAL AND INTEGRAL CALCULUS. J, Jo *Ydxdy = J b [J o xYdy)dx =i'(¥)*=^'-»' >• That is, as dy is written last, the ^-integration is taken first. The limits of the first integration are often functions of the second variable. For example, /a /» Vx n a I dx dy- I ( Vx )dx = fa 1 . Jo Jo As another example, rdrd9 = J Qdr = A b\ 233. A Triple Integral is the indicated operations of three successive integrations, for which the notation is similar to that of double integrals. Thus, £ a £fJ x y zdxd y ih =X\Jo\fJ x y zdz ) d !i'] dx - = £ a \J, X{ix * yi)dy ~] dx = J? ww* = ^ a °- EXAMPLES. Find the following : L jf jf>° + y,)dx d,j - ir,b{a ' + 6> >- ~- / / {f+f)lxdy. \na\ Jo Jo 3. 2a l 'a I rVa?-ax 1 clx clz ^ Vax - -X* x* 4- ?/ 2 a dxdy 4. •VOax-x* / clz. INTEGRATION AS A SUMMATION OF ELEMENTS. 233 lna\ AREAS OF SURFACES DETERMINED BY DOUBLE INTEGRATION. 234. Plane Surfaces — (a) Rectangular Co-ordinates. In the formula u = I y dx, Art. 226, we may make y = I dy, which gives u = I j dxdy (J) /r 2 F > '—"- — >»^^ v/^ D >v / q P \ n \ / m B C Fig. 50. Art. 228, we may make - = / rdr, and write u = I I rdddr. . . • • (K) 234 DIFFERENTIAL AND INTEGRAL CALCULUS. As illustrative examples let us employ (J) and (K) in finding the area of a circle whose radius is a. Let (x, y) be the co-ordinates of the point m, and (x -f- dx, y -\- dy) of the point p, then mnpq = dxdy. Kegarding x as constant, we have v=bp BGDP = 2 [mnpq] y=o \/2ax-x* /» \2ax-x* = dx I dy = \2ax — x 2 dx. Again, since V 2ax — x*dx is the differential value of BP' with respect to dx, we have 2a p- a _ ? [BP'] = / V2 ^0 ax x 2 dx = in a? = area of OKP. (b) Let (r, 0) be the co-ordinates of m, and (r -\- dr, -\- dd) of //, then mn = rfr, wg = w?0, and rdddr ( = area of mnpq) is J 3 / Fig. 51. the differential value of the element mnp'q' with respect to dr Therefore, regarding 6 as constant, we have 2 [mnp'q'] = I rdddr = 2a 2 cos 2 Odd = 0PP x '. o * INTEGRATION AS A SUMMATION OF ELEMENTS. 235 Again, since the area of OPP/ is the differential value of the element OPP', with respect to d6, we have 2*[0PP'] = f 2a" cos* Odd = ina\ which is one half the area of the circle. EXAMPLES. 1. Find the area (1) of a rectangle by double integration; (2) of a triangle. 2. Find the area between the parabola «/ 2 = ax and the circle y* = 2ax — x\ J7ta* 2a 2 \ 3. Find by double integration the entire area of the cardioid r = a(l — cos 0). Zna 2 ~2~' 4. In a similar manner find the entire area of the Lemniscate r* = a 2 cos2d. a\ 5. Find the whole area between the curve xy' 1 = 4a 2 (2a — x) and its asymptote. ^na?. 236. Surfaces in General.— To find the area (= S) of a surface whose equation is f(x, y y z) = 0. Let (x, y, z) be the co-ordinates of any point P of the surface, and (x -f- dx, y -f- dy, z + dz) the co-ordinates of a second point Q very near the first (Fig. 52). Draw planes through P and Q parallel to the planes XZ and YZ. These planes will intercept a curved quadrilateral PQ on the surface; its projection pq, a rectangle, on the plane of XY] and a parallelogram p'q' , not shown in the figure, on the tangent plane at P, of which pq is the projection. The area of p'q' — dS, since it is the dif- ferential value of PQ (= AS) with respect to dx and dy. The projection of p'q' on XY is dxdyx similarly the projec- tions of p'q' on XZ and YZ are dxdz and dydz; hence, denoting 236 DIFFERENTIAL AND INTEGRAL CALCULUS. the angles between the plane otp'q' and XY, XZ and YZ by a, fi and y, respectively, we have cos adS = dxdy, (1) cos fidS = dxdz, (2) cos ydS = dydz (3) Fig. 52. Squaring (1), (2), (3), and adding, remembering that cos 2 a -f cos 2 /3 -f cos 2 y = 1, we have hence, (dS)' = dx'dy* + dx*dz> + difdz*; INTEGRATION AS A SUMMATION OF ELEMENTS. 237 EXAMPLES. 1. Find the area of one eighth of the surface of the sphere x- + y 2 + 2 a = a\ „ dz x dz v Here -j- = , -7- = - -. dx z dy z "^ £&* "^ Jo Jo Va 2 - x 2 - y 2 ' 2 " 2. The two cylinders x 2 -f z 2 — a 2 and x 2 -\- y 2 = a 2 intersect at right angles; find the surface of the one intercepted by the other. 8a 2 . Here z = Va 2 — x 2 . and for one eighth of the required sur- face the ^/-limits are and Va 2 — x 2 , and the ^-limits and a. 3. A sphere whose radius is a is cut by a right circular cylinder, the radius of whose base is — , and one of whose edges passes through the centre of the sphere; find the area of the surface of the sphere intercepted by the cylinder. 2a 2 (n — 2). Take x 2 + y 2 -\- z 2 = a 2 for the sphere, and x 2 -f y* = ax for the cylinder, then z = Va 2 — y 2 — x 2 , and for one fourth of the 238 DIFFERENTIAL AND INTEGRAL CALCULUS. required surface the limits of y and x are 0, Vax — x 2 , and 0, a, respectively. 4. In the preceding example find the surface of the cylinder intercepted by the sphere. (See Ex. 3, Art. 233.) 4a 2 . 5. Find the area of the portion of the surface of the sphere x % -j- y" 2 -f- z 2 = 2ay lying within the paraboloid y = mx 2 + nz 2 > 27ta Vmn 237. To find the volume of a solid bounded by a surface whose equation is f(x, y, z) = 0. Fig. 53. VOLUMES OF SOLIDS DETERMINED BY TRIPLE INTEGRATION. Let v = the indefinite volume expressed by the product of x, y, z; then v = xyz, which may be written v = I I I dxdy dz, which becomes definite when the integrations are taken between certain limits, and we will now give the geometrical interpreta- tion of the formula, step by step. INTEGRATION AS A SUMMATION OF ELEMENTS. 239 Let (x, y, z) be the co-ordinates of the point P, and (x -f- dx, y -f dy, z -f- dz) be the co-ordinates of the point Q; then PQ = dxdy dz. (a) Kegarding x and y as constant and integrating between the 2-limits and id, we have 2 M [PQ] = J^dxdydz = (id)dxdy=M. (b) The volume of bh is the differential value of the element is' with respect to dy, hence nam 2™ [is'] = J o (id)dx dy = (afm)dx, which is the volume of the cylindrical segment afm-a'. (c) The volume of afm-a' is the differential value of the element ar with respect to dx; hence ^ o \ar] = I (afm)dx = volume of OBC-A. pOA nam r*i& J o J o J o dxdydz — volume of OBC-A. . . (M) Cor. I. The limits of y and z are found thus: id is the posi- tive result of solving the equation f(x, y, z) = for z, am is the positive result of solving f(x, y,0) = for y, and OA is the positive result of solving f(x, 0, 0) = for x. EXAMPLES. 1. Find the volume of one eighth of the ellipsoid a* ^ b* ^ c* I x* ?/ 2 \* The limits of z in this case are and id = ell i* — fr) 5 240 DIFFERENTIAL AND INTEGRAL CALCULUS. the limits of y are and am = bi 1 -j ; and the limits of x are and a. Therefore the required volume is J. V a2 r c * l a i ; j2 / dx dy dz = 6 2. Find the volume of the solid contained between (a) the paraboloid of revolution, x* -\- y 2 = az, (b) the cylinder, x 1 -f y' = 2ax, (c) and the plane, 2 = 0. 37ra a 2 " # 2 4- ?/ 2 The 2-limits are and — J , the y-limits are and V2ax — x 1 , and the ^-limits are and 2a, for one half of the required volume. 3. Find the volume cut from a sphere whose radius is a by a right circular cylinder whose radius is b, and whose axis passes through the centre of the sphere. irf . 3X s F x(« ! - («' - *)')• 4. Find the entire volume bounded by the surface whose equation is x* -f- y* + s? = a 1 . 4?rfl 3 35 ' 5. Find the volume of the conoid bounded by the surface z*x* + a*y* = cV, and the planes x = and x = a. %itac\ APPLICATION TO MECHANICS. 238. "Work is said to be done when a body moves through space in opposition to resistance. A horse in drawing a cart or a plough does a certain amount of work, which depends on the resistance and the distance traversed. The force which the horse exerts, and the distance through which he moves, may be regarded as the two elements of the work done. If r lbs. is the constant resistance or force, and x feet the effective distance INTEGRATION AS A SUMMATION OF ELEMENTS, 241 through which the body moves, rx units of work will be done. By effective distance is meant the distance measured in the same direction as that in which the force is acting. Thus, when the resistance is constant, the amount of work may be represented by the area of a rectangle whose base is the distance (x) and whose altitude is the resistance (r). If the resistance or force is a variable dependent on the dis- tance x, it may be represented hjf'(x), in which case the amount of work may be found by taking the sum of its elements, thus: In Fig. 43, if the force fix) (= BP) act through the small effective distance h (= BC), the work done will be in excess of f'(x)h (= BCDP) only by the acceleration of work {PDP f ) during that interval. Hence, /' (x)dx is the differential value of the zth element of work. Therefore the quantity of work between the. limits x = x x and x = # 2 is, viz. : 2lWov l ™l^y\x)l l -\=f x y'{x)dx, . (N) in which the effective distance is z 2 — x x . Cor. I. Effective distance, resistance, and work, and effective distance, force, and energy, bear the same relation to one another as the abscissa, ordinate, and area of a plane curve referred to rectangular co-ordinates, respectively. Example. Let it be required to compute the quantity of work necessary to compress the spiral spring of the common spring- balance to any given degree, say from AB to DB.* Let the resistance ( = /'(#)) vary directly as the degree of conrpression, and denote the distance AD' by x\ then will f'{x) == mx, Fig. 54. where m is the resistance of the spring when the balance Bartlett's Analytical Mechanics, page 39. 242 DIFFERENTIAL AND INTEGRAL CALCULUS. is compressed through the distance unity. Substituting in (N), making x l = and x a — AD, we have the work = / mx dx = — 1- G = -kmx n o Cor. I. If m = 10 pounds and x 2 — 3 ft., then will the work = 45 units of work; that is, the quantity of work will be equal to that required to raise 45 pounds through a vertical height of one foot. CENTRE OF GRAVITY. 239. The bodies here considered are supposed to be of uni- form density; that is, equal quantities of a body have equal weights. The Centre of Gravity of a body is a point so situated that the force of gravity produces no tendency in the body to rotate about any axis passing through this point. The Moment of any element or particle of a body with reference to any horizontal axis is the product of the magnitude or weight of the element by the horizontal distance of its centre of gravity from the axis, and measures the tendency of the element, under the influence of gravity, to produce rotation about the axis. The moment of the body itself is the sum of the moments of its elements. If the axis of reference passes through the centre of gravity of the body, the moment of the body must be zero, otherwise the moments of the elements would not neutralize one another, and the body would rotate. 240. To find the centre of gravity of a plane area. In Fig. 43, suppose the plane curve placed in a horizontal position, and let A = the area of EGQF, % — OB, y = BP, x i — OE, x 2 — OG. Also let {x',y') be the centre of gravity of A, and (x -\- a, y -j- /3) be the centre of gravity of the rect- angle BCDP (= yh). Evidently the limit of a, as BC or h approaches 0, is 0. INTEGRATION AS A SUMMATION OF ELEMENTS. 243 The moment of BCDP with respect to an axis passing through (x' ', y') and parallel to the axis of y is (x -j- a — x')y]i, which is the measure of the tendency of this rectangle [BCDP) to produce rotation about the given axis, and therefore the tendency of all the similar rectangles to produce rotation is ^ \x -\- a — x']yJi. The smaller the rectangles the nearer their sum comes to the whole area of the curve, and therefore the tendency of A to rotate about the given axis is , . ^ X \x + <* — x')yli —I i x — x')ydx: h or a _ x l v ' . }J «4, but as the axis of reference passes through the centre of gravity of A, this must equal zero. / {x — x')ydx — I xydx — x' I ydx = 0. xydx -^- / ydx = / xydx ~ A. (P) In like manner we find y' f X 'fdx + A (Q) 241. To find the centre of gravity of a plane curve- In Fig. 44, suppose the plane curve PQ (= s) placed in a horizontal position; let x = OB, y = BP, x 2 — OG, and h = BC; also let (#', y') be the centre of gravity of s, and (x -f- a, y -f- fi) the centre of gravity of the tangent The limit of a, as h approaches 0, is evidently 0. 244 DIFFERENTIAL AND INTEGRAL CALCULUS. The moment of Pt with respect to an axis passing through (x' t y') and parallel to the axis of y is (%-\- 10. Find the centre of gravity of the paraboloid generated by revolving y 2 = kmx about the axis of x. (See formula (T).) x' — \x. 11. Find the centre of gravity of the segment of a sphere- generated by revolving y 2 = 2rx — x 2 about the axis of x. , _ x(8r — 3x) 4(3r — x) ' When x = r, %' = §r. 12. A semi-ellipsoid is formed by the revolution of a semi- ellipse about its major axis; find the distance of the centre of gravity of the solid from the centre of the ellipse. x' = § a. 13. Find the centre of gravity of the convex surface of the cone generated by revolving the line y = mx about the axis of x. (See formula (U).) x f = §&. 14. Find the centre of gravity of the surface of a spherical segment whose altitude is x. x f = \x. 15. Find the centre of gravity of the surface of the parabo- loid generated by revolving y 2 = Amx about the axis of x. , 1 (3z - 2m) (x + mf + 2w* . 5 (x + ra) 1 — wi 1 APPENDIX. A,. Differentiable Functions. A function, y = f(x), is Ay said to be differentiable when -~ approaches a definite limit as Ax approaches zero. Thus, y = Vx is differentiable, since (Art. 10, ex. 5) 4y _ 1 Ax |V + A + Vx' . approaches the definite limit, — — , as h approaches zero, x f 4irX being any particular or definite value of x from which li is estimated. All ordinary continuous functions are differentiable, but this does not follow from the mere fact that the functions are con- tinuous, for there are functions which are continuous and yet have no differential coefficients.* Functions of this kind, how- ever, are of such rare occurrence that the distinction between continuity and differentiability is seldom made in works on the Differential Calculus. That is, every function is regarded as continuous and differentiable between certain limits. Ati The limit (m x ) of -p in any particular case can often be conveniently determined by assuming that Ay = mji + mji 2 , * See Harkness and Morley's Theory of Functions. 247 248 APPENDIX, which is true of all differentiate functions of a single variable, and then finding the value of m l9 as in A 4 . The general values of m\ and m 2 , and the exact conditions under which Ay = mji -j- mji' 2 holds, are given in A 7 . A 2 . Another Illustration of the Formula Ay = mji + mji 2 . Suppose that a moving body has traversed a distance (s) in the time t, and that the value of s in terms of t is *=/(0 (i) Suppose we wish to find the actual velocity (v x ) at the end of the time /,. Let At, an increment of t estimated from t l9 be any arbitrary period immediately succeeding the end of the time t l9 then the distance traversed by the body in that period Avill be the corresponding increment of s, viz., As = f(t + At) - f{t) = m x At + m,(At)\ . . (2) The mean velocity (v) of a moving body, during any period of time, is the quotient obtained by dividing the distance traversed by the body by the length of the period. Therefore the mean velocity during the period At is ^ = m t + m 2 At =.v (3) Now this mean velocity evidently approaches the actual velocity v x as At approaches 0, indefinitely. Hence taking the limit of (3), we have limit of —r- = > m 1 = v x . As That is, the limit of —r, or m 1 , is the actual velocity of the body at the end of the time t x , and hence m x At is what As would have been had it varied as At or had the actual velocity i\ re- mained constant, and m 2 /i 2 is the acceleration of s during the period At, APPENDIX. 249 A 3 . The Differentials of Independent Variables are, in general, Variables. In differentiating y = f(x) successively dx is usually regarded as a constant; that is, as having the same value for all values of x. " This hypothesis/' say Eice and John- son,* "greatly simplifies the expressions for the second and higher differentials of functions of x, inasmuch as it is evidently equivalent to making all differentials of x higher than the first vanish." Again, "A differential of the second order or of a higher order," says Byerly,f " has been defined by the aid of a derivative, which always implies the distinction between func- tion and variable, and on the hypothesis of an important dif- ference in the natures of the increments of function and vari- able; namely, that the increment of the independent variable is a constant magnitude, and that, consequently, its derivative and differential are zero." The impressions which these and similar statements in other excellent works are likely to make on the mind of the student are (a) that all independent variables vary uniformly, and (b) that they must vary in this manner in order that the differen- tials of their differentials shall be zero. That an independent variable may vary uniformly, as in Eate of Change, is granted; but differentials in general are variables whose limits are zero. Indeed one of the most important and essential properties of the differential of an independent vari- able is its independent variability. The imposition of any con- dition on a group of variables by w T hich they may be expressed in terms of one another at once destroys the independence of the variables, and this is the case of variables under the hypothesis of uniform variation, or rate of change. Thus, let u =f(x, y), and let us suppose x and y to vary simultaneously and uniformly; then dx = mdt and dy = m'dt; whence x = mt -f C and y — m't + C. Eliminating t and solving for y, we have y = (x). Hence the supposition of uniform* change renders the hypothesis of more than one in- * Dif . Calculus, Art. 79. f Dif . Calculus, Art. 204. 250 APPENDIX. dependent variable impossible. Therefore, if independent variables (which are supposed to vary simultaneously) must vary uniformly in order that their higher differentials shall be zero, the successive differentials of u = f(y, x, z) can be obtained only by destroying the independence of all the variables x 9 y, z except one. Hence, in general, if the differential of dx, with respect to x, is zero, it is due to the' fact that dx is a variable which is independent of x. A 4 . To differentiate -a v and log a v independently of Art. 78. Let y = a v , where v is a function of x. Increasing x by h, etc., and assuming that a v is differentiable (AJ, we have Ay , (a^ — 1\ , a Av \ Av J * a Taking the limits, remembering that m^Av vanishes with h and that a v is constant with respect to li, we have where m 1 and a v are definite quantities. Therefore the limit of - must be a definite quantity (m f say), not zero, and de- ZjV pendent solely on the base a, since it is evidently independent of x and h. ~j — a v m r , or dy = tfm'dv (1) Cor. I. Since m' depends on the base and the base is arbi- trary, we may suppose the base to have such a value (say e) that APPENDIX. 251 Coe. II. To find the value of m' in (1), let a v = e u (2) Differentiating (2), a v m'dv — e u du (3) rjii. From (2) and (3), m' = ~: .;..... (4) Taking log e of (2), v log e a = u (5) Differentiating (5), dv log e a = du (6) Whence log e a = — (7) Equating (4) and (7), m* — log e a. (8) d{a v ) = a v log, a dv (9) Cok. III. To differentiate y — log a v, we write it under the form of a y = v (10) Differentiating (10), a v log c ady — dv; whence dv = , .or log a e — (11) J log e a v & v ' A 5 . A Rigorous Proof of Taylor's Formula. In what fol- lows, the function f(y) and its n successive derivatives are sup- posed to be differentiable, and finite and continuous between the limits y and y -f x. Lemma. If F(z) is continuous bet wen z = a and z = y, and if F(a) = F{y) = 0, then F\z), if continuous, must equal zero for some value of z between a and y. For, as z changes from a to y, F(z) passes from to 0, that is, F(z) increases and then decreases, or vice versa; hence F'{z) must change from -J- to — or from — to -f- , and therefore, since it is continuous, pass through 0. 252 APPENDIX. In what follows 6 will represent a positive proper fraction; that is, < 6 < 1. Hence, < dx < x, and, by the lemma, F\y + B(a - y)] = 0. Under the given hypotheses, we have (A,) f n -\y + x) =-f n -\y) + m x x + m